\displaystyle \text{Solve the following problems based on differential equations:}

\displaystyle \textbf{Question 1: }\text{The surface area of a balloon being inflated changes at a rate} \\ \text{proportional to time } t. \text{ If initially its radius is }1\text{ unit and } \text{after }3\text{ seconds it is }2\text{ units, } \\ \text{find the radius after time }t.
\displaystyle \text{Answer:}
\displaystyle \text{Let } r \text{ be the radius and } S \text{ be the surface area of the balloon at time } t
\displaystyle S=4\pi r^{2}
\displaystyle \Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt} \qquad (1)
\displaystyle \text{Given: } \frac{dS}{dt}\propto t
\displaystyle \Rightarrow \frac{dS}{dt}=kt
\displaystyle \text{Putting } \frac{dS}{dt}=kt \text{ in (1), we get}
\displaystyle kt=8\pi r\frac{dr}{dt}
\displaystyle \Rightarrow kt\,dt=8\pi r\,dr
\displaystyle \text{Integrating both sides,}
\displaystyle \int kt\,dt=\int 8\pi r\,dr
\displaystyle \Rightarrow \frac{kt^{2}}{2}=4\pi r^{2}+C \qquad (2)
\displaystyle \text{At } t=0,\ r=1
\displaystyle \Rightarrow 0=4\pi(1)^{2}+C
\displaystyle \Rightarrow C=-4\pi
\displaystyle \text{Also, at } t=3,\ r=2
\displaystyle \Rightarrow \frac{9k}{2}=4\pi(4)-4\pi
\displaystyle \Rightarrow \frac{9k}{2}=12\pi
\displaystyle \Rightarrow k=\frac{8\pi}{3}
\displaystyle \text{Substituting values of } k \text{ and } C \text{ in (2),}
\displaystyle \frac{4\pi t^{2}}{3}=4\pi r^{2}-4\pi
\displaystyle \Rightarrow \frac{t^{2}}{3}=r^{2}-1
\displaystyle \Rightarrow r^{2}=1+\frac{t^{2}}{3}
\displaystyle \Rightarrow r=\sqrt{1+\frac{t^{2}}{3}}

\displaystyle \textbf{Question 2: }\text{A population grows at the rate of }5\%\text{ per year. How long does it} \\ \text{take for the population to double?}
\displaystyle \text{Answer:}
\displaystyle \text{Let } P_{0} \text{ be the initial population and } P \text{ be the population at time } t
\displaystyle \frac{dP}{dt}=\frac{5P}{100}
\displaystyle \Rightarrow \frac{dP}{dt}=0.05P
\displaystyle \Rightarrow \frac{dP}{P}=0.05\,dt
\displaystyle \text{Integrating both sides with respect to } t,
\displaystyle \int\frac{dP}{P}=\int 0.05\,dt
\displaystyle \Rightarrow \log P=0.05t+C
\displaystyle \text{Now, } P=P_{0} \text{ at } t=0
\displaystyle \Rightarrow \log P_{0}=C
\displaystyle \text{Putting the value of } C,
\displaystyle \log P=0.05t+\log P_{0}
\displaystyle \Rightarrow \log\left(\frac{P}{P_{0}}\right)=0.05t
\displaystyle \text{To find the time when the population will double, take } P=2P_{0}
\displaystyle \Rightarrow \log 2=0.05t
\displaystyle \Rightarrow t=\frac{\log 2}{0.05}
\displaystyle \Rightarrow t=20\log 2 \text{ years}

\displaystyle \textbf{Question 3: }\text{The rate of growth of a population is proportional to the number present.} \\ \text{If the population of a city doubled in the past }25\text{ years, and the present population is } \\ 100000,\text{ when will the city have a population of }500000?\\ \text{[Given } \log_e 5=1.609,\ \log_e 2=0.6931\text{].}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original population be } N \text{ and the population at any time } t \text{ be } P
\displaystyle \text{Given: } \frac{dP}{dt}=aP
\displaystyle \Rightarrow \frac{dP}{P}=a\,dt
\displaystyle \text{Integrating,}
\displaystyle \int\frac{dP}{P}=\int a\,dt
\displaystyle \Rightarrow \log|P|=at+C \qquad (1)
\displaystyle \text{Now, } P=N \text{ at } t=0
\displaystyle \Rightarrow \log|N|=C
\displaystyle \text{Putting } C=\log|N| \text{ in (1),}
\displaystyle \log|P|=at+\log|N|
\displaystyle \Rightarrow \log\left(\frac{P}{N}\right)=at \qquad (2)
\displaystyle \text{According to the question, } \log\left(\frac{2N}{N}\right)=25a
\displaystyle \Rightarrow \log 2=25a
\displaystyle \Rightarrow a=\frac{\log 2}{25}
\displaystyle \text{For } P=500000 \text{ and } N=100000,
\displaystyle \log\left(\frac{500000}{100000}\right)=at
\displaystyle \Rightarrow \log 5=\frac{\log 2}{25}t
\displaystyle \Rightarrow t=\frac{25\log 5}{\log 2}= \text{approx } 58

\displaystyle \textbf{Question 4: }\text{In a culture, the bacteria count is }100000.\text{ The number increased by }\\ 10\%\text{ in } 2\text{ hours. In how many hours will the count reach }200000,\text{ if the rate of growth is} \\ \text{proportion to the number present?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the bacteria count at any time } t \text{ be } N
\displaystyle \text{Given: } \frac{dN}{dt}=\lambda N
\displaystyle \Rightarrow \frac{dN}{N}=\lambda\,dt
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{dN}{N}=\int \lambda\,dt
\displaystyle \Rightarrow \log N=\lambda t+\log C \qquad (1)
\displaystyle \text{Initially, when } t=0,\ N=100000
\displaystyle \Rightarrow \log 100000=\log C
\displaystyle \Rightarrow C=100000
\displaystyle \text{After 2 hours, number increased by } 10\%
\displaystyle \Rightarrow N=110000 \text{ when } t=2
\displaystyle \text{Putting } t=2,\ N=110000 \text{ in (1),}
\displaystyle \log 110000=2\lambda+\log 100000
\displaystyle \Rightarrow 2\lambda=\log\left(\frac{110000}{100000}\right)
\displaystyle \Rightarrow \lambda=\frac{1}{2}\log\left(\frac{11}{10}\right)
\displaystyle \text{Substituting values of } C \text{ and } \lambda \text{ in (1),}
\displaystyle \log N=\frac{t}{2}\log\left(\frac{11}{10}\right)+\log 100000 \qquad (2)
\displaystyle \text{Let } t=T \text{ when } N=200000
\displaystyle \text{Putting in (2),}
\displaystyle \log 200000=\frac{T}{2}\log\left(\frac{11}{10}\right)+\log 100000
\displaystyle \Rightarrow \log 2=\frac{T}{2}\log\left(\frac{11}{10}\right)
\displaystyle \Rightarrow T=\frac{2\log 2}{\log\left(\frac{11}{10}\right)}

\displaystyle \textbf{Question 5: }\text{If the interest is compounded continuously at }6\%\text{ per annum, how } \\ \text{much worth Rs }1000\text{ will be after }10\text{ years? How long will it take to } \\ \text{double Rs }1000? \text{[Given }e^{0.6}=1.822\text{].}
\displaystyle \text{Answer:}
\displaystyle \text{Let } P_{0} \text{ be the initial amount and } P \text{ be the amount at time } t
\displaystyle \frac{dP}{dt}=\frac{6P}{100}
\displaystyle \Rightarrow \frac{dP}{dt}=0.06P
\displaystyle \Rightarrow \frac{dP}{P}=0.06\,dt
\displaystyle \text{Integrating both sides with respect to } t,
\displaystyle \int\frac{dP}{P}=\int 0.06\,dt
\displaystyle \Rightarrow \log P=0.06t+C
\displaystyle \text{Now, } P=P_{0} \text{ at } t=0
\displaystyle \Rightarrow \log P_{0}=C
\displaystyle \text{Putting the value of } C,
\displaystyle \log P=0.06t+\log P_{0}
\displaystyle \Rightarrow \log\left(\frac{P}{P_{0}}\right)=0.06t
\displaystyle \Rightarrow \frac{P}{P_{0}}=e^{0.06t}
\displaystyle \text{To find the amount after 10 years,}
\displaystyle \frac{P}{P_{0}}=e^{0.6}
\displaystyle \Rightarrow P=e^{0.6}P_{0}
\displaystyle \Rightarrow P\approx 1.822P_{0}
\displaystyle \text{If } P_{0}=1000,
\displaystyle P\approx 1822\ \text{Rs}
\displaystyle \text{To find the time for the amount to double, take } P=2P_{0}
\displaystyle \Rightarrow \log 2=0.06t
\displaystyle \Rightarrow t=\frac{\log 2}{0.06}
\displaystyle \Rightarrow t\approx 11.55\ \text{years}

\displaystyle \textbf{Question 6: }\text{The rate of increase in the number of bacteria in a certain bacteria } \\ \text{culture is proportional to the number present. Given the number triples in }5\text{ hours, } \\ \text{find how many bacteria will be present after }10\text{ hours. Also find the time necessary } \\ \text{for the number of bacteria to be }10\text{ times the number of initial present.}\\ \text{[Given } \log_e 3=1.0986,\ e^{2.1972}=9\text{].}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original count of bacteria be } N \text{ and the count at any time } t \text{ be } P
\displaystyle \text{Given: } \frac{dP}{dt}=aP
\displaystyle \Rightarrow \frac{dP}{P}=a\,dt
\displaystyle \text{Integrating,}
\displaystyle \int\frac{dP}{P}=\int a\,dt
\displaystyle \Rightarrow \log|P|=at+C \qquad (1)
\displaystyle \text{Now, } P=N \text{ at } t=0
\displaystyle \Rightarrow \log|N|=C
\displaystyle \text{Putting } C=\log|N| \text{ in (1),}
\displaystyle \log|P|=at+\log|N|
\displaystyle \Rightarrow \log\left(\frac{P}{N}\right)=at \qquad (2)
\displaystyle \text{According to the question, } \log\left(\frac{3N}{N}\right)=5a
\displaystyle \Rightarrow \log 3=5a
\displaystyle \Rightarrow a=\frac{1}{5}\log 3
\displaystyle \text{Substituting in (2),}
\displaystyle \log\left(\frac{P}{N}\right)=\frac{t}{5}\log 3 \qquad (3)
\displaystyle \text{Putting } t=10 \text{ in (3),}
\displaystyle \log\left(\frac{P}{N}\right)=2\log 3
\displaystyle \Rightarrow \frac{P}{N}=e^{2\log 3}=9
\displaystyle \Rightarrow P=9N
\displaystyle \text{To find the time when } P=10N,
\displaystyle \log 10=\frac{t}{5}\log 3
\displaystyle \Rightarrow t=\frac{5\log 10}{\log 3}

\displaystyle \textbf{Question 7: }~\text{The population of a city increases at a rate proportional to the number } \\ \text{of inhabitants present at any time }t. \text{ If the population of the city was }200000\text{ in } \\ 1990\text{ and }250000\text{ in }2000,\text{ what will be the population in }2010?
\displaystyle \text{Answer:}
\displaystyle \text{Let the population at any time } t \text{ be } P
\displaystyle \text{Given: } \frac{dP}{dt}=\beta P
\displaystyle \Rightarrow \frac{dP}{P}=\beta\,dt
\displaystyle \text{Integrating,}
\displaystyle \int\frac{dP}{P}=\int \beta\,dt
\displaystyle \Rightarrow \log|P|=\beta t+\log C \qquad (1)
\displaystyle \text{Now, at } t=1990,\ P=200000
\displaystyle \Rightarrow \log 200000=1990\beta+\log C \qquad (2)
\displaystyle \text{At } t=2000,\ P=250000
\displaystyle \Rightarrow \log 250000=2000\beta+\log C \qquad (3)
\displaystyle \text{Subtracting (3) from (2),}
\displaystyle \log 200000-\log 250000=-10\beta
\displaystyle \Rightarrow \beta=\frac{1}{10}\log\left(\frac{5}{4}\right)
\displaystyle \text{Putting } \beta=\frac{1}{10}\log\left(\frac{5}{4}\right) \text{ in (2),}
\displaystyle \log 200000=1990\cdot\frac{1}{10}\log\left(\frac{5}{4}\right)+\log C
\displaystyle \Rightarrow \log C=\log 200000-199\log\left(\frac{5}{4}\right)
\displaystyle \text{Putting values in (1) for } t=2010,
\displaystyle \log|P|=\frac{2010}{10}\log\left(\frac{5}{4}\right)+\log 200000-199\log\left(\frac{5}{4}\right)
\displaystyle \Rightarrow \log|P|=201\log\left(\frac{5}{4}\right)+\log 200000-199\log\left(\frac{5}{4}\right)
\displaystyle \Rightarrow \log|P|=\log\left(\left(\frac{5}{4}\right)^{2}\right)+\log 200000
\displaystyle \Rightarrow \log|P|=\log\left(\frac{25}{16}\cdot 200000\right)
\displaystyle \Rightarrow \log|P|=\log 312500
\displaystyle \Rightarrow P=312500

\displaystyle \textbf{Question 8: }~\text{If the marginal cost of manufacturing a certain item is given by }C'(x)=\frac{dC}{dx}=2+0.15x,\text{ find the total cost function }C(x),\text{ given that }C(0)=100.
\displaystyle \text{Answer:}
\displaystyle \text{We have, } \frac{dC}{dx}=2+0.15x
\displaystyle \Rightarrow dC=(2+0.15x)\,dx
\displaystyle \text{Integrating both sides with respect to } x,
\displaystyle C=\int(2+0.15x)\,dx
\displaystyle =2x+\frac{0.15}{2}x^{2}+K \qquad (1)
\displaystyle \text{At } C(0)=100,
\displaystyle 100=2(0)+\frac{0.15}{2}(0)^{2}+K
\displaystyle \Rightarrow K=100
\displaystyle \text{Putting the value of } K \text{ in (1),}
\displaystyle C=2x+\frac{0.15}{2}x^{2}+100
\displaystyle \Rightarrow C=0.075x^{2}+2x+100

\displaystyle \textbf{Question 9: }~\text{A bank pays interest by continuous compounding, that is, by treating the } \\ \text{interest rate as the instantaneous rate of change of principal. Suppose in an account } \\ \text{interest accrues at }8\%\text{ per year, compounded continuously. Calculate the percentage } \\ \text{increase in such an account over one year.}\\ \text{[Take }e^{0.08}=1.0833\text{].}
\displaystyle \text{Answer:}
\displaystyle \text{Let } P_{0} \text{ be the initial amount and } P \text{ be the amount at time } t
\displaystyle \frac{dP}{dt}=\frac{8P}{100}
\displaystyle \Rightarrow \frac{dP}{dt}=\frac{2P}{25}
\displaystyle \Rightarrow \frac{dP}{P}=\frac{2}{25}\,dt
\displaystyle \text{Integrating both sides with respect to } t,
\displaystyle \int\frac{dP}{P}=\int\frac{2}{25}\,dt
\displaystyle \Rightarrow \log P=\frac{2}{25}t+C \qquad (1)
\displaystyle \text{Now, } P=P_{0} \text{ at } t=0
\displaystyle \Rightarrow \log P_{0}=C
\displaystyle \text{Putting the value of } C,
\displaystyle \log P=\frac{2}{25}t+\log P_{0}
\displaystyle \Rightarrow \log\left(\frac{P}{P_{0}}\right)=\frac{2}{25}t
\displaystyle \Rightarrow \frac{P}{P_{0}}=e^{\frac{2}{25}t}
\displaystyle \text{To find the amount after 1 year, put } t=1
\displaystyle \Rightarrow \frac{P}{P_{0}}=e^{\frac{2}{25}}
\displaystyle \Rightarrow \frac{P}{P_{0}}\approx 1.0833
\displaystyle \Rightarrow P\approx 1.0833P_{0}
\displaystyle \text{Percentage increase }=\frac{P-P_{0}}{P_{0}}\times 100
\displaystyle =0.0833\times 100
\displaystyle \Rightarrow 8.33\%

\displaystyle \textbf{Question 10: }~\text{In a simple circuit of resistance }R,\text{ self inductance }L\text{ and voltage } \\ E,\text{ the current }i\text{ at any time }t\text{ is given by }L\frac{di}{dt}+Ri=E.\text{ If } E\text{ is constant } \\ \text{and initially no current passes through the circuit, prove that }  i=\frac{E}{R}\left(1-e^{-(R/L)t}\right).
\displaystyle \text{Answer:}
\displaystyle \text{We have, } L\frac{di}{dt}+Ri=E
\displaystyle \Rightarrow \frac{di}{dt}+\frac{R}{L}i=\frac{E}{L} \qquad (1)
\displaystyle \text{I.F.}=e^{\int\frac{R}{L}\,dt}=e^{\frac{R}{L}t}
\displaystyle \text{Multiplying both sides of (1) by I.F.}=e^{\frac{R}{L}t}
\displaystyle e^{\frac{R}{L}t}\left(\frac{di}{dt}+\frac{R}{L}i\right)=e^{\frac{R}{L}t}\frac{E}{L}
\displaystyle \Rightarrow \frac{d}{dt}\left(ie^{\frac{R}{L}t}\right)=\frac{E}{L}e^{\frac{R}{L}t}
\displaystyle \text{Integrating both sides with respect to } t,
\displaystyle ie^{\frac{R}{L}t}=\frac{E}{L}\int e^{\frac{R}{L}t}\,dt+C
\displaystyle =\frac{E}{L}\cdot\frac{L}{R}e^{\frac{R}{L}t}+C
\displaystyle =\frac{E}{R}e^{\frac{R}{L}t}+C \qquad (2)
\displaystyle \text{Now, } i=0 \text{ at } t=0
\displaystyle \Rightarrow 0=\frac{E}{R}+C
\displaystyle \Rightarrow C=-\frac{E}{R}
\displaystyle \text{Putting the value of } C \text{ in (2),}
\displaystyle ie^{\frac{R}{L}t}=\frac{E}{R}\left(e^{\frac{R}{L}t}-1\right)
\displaystyle \Rightarrow i=\frac{E}{R}\left(1-e^{-\frac{R}{L}t}\right)

\displaystyle \textbf{Question 11: }~\text{The decay rate of radium at any time }t\text{ is proportional to its } \\ \text{mass at that time. Find the time when the mass will be halved of its initial mass.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the initial amount of radium be } N \text{ and the amount present at time } t \text{ be } P
\displaystyle \text{Given: } \frac{dP}{dt}\propto P
\displaystyle \Rightarrow \frac{dP}{dt}=-aP,\ a>0
\displaystyle \Rightarrow \frac{dP}{P}=-a\,dt
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{dP}{P}=\int -a\,dt
\displaystyle \Rightarrow \log|P|=-at+C \qquad (1)
\displaystyle \text{Now, } P=N \text{ at } t=0
\displaystyle \Rightarrow \log|N|=C
\displaystyle \text{Putting } C=\log|N| \text{ in (1),}
\displaystyle \log|P|=-at+\log|N|
\displaystyle \Rightarrow \log\left(\frac{N}{P}\right)=at
\displaystyle \text{According to the question, } P=\frac{N}{2}
\displaystyle \Rightarrow \log\left(\frac{N}{N/2}\right)=at
\displaystyle \Rightarrow \log 2=at
\displaystyle \Rightarrow t=\frac{1}{a}\log 2

\displaystyle \textbf{Question 12: }~\text{Experiments show that radium disintegrates at a rate proportional to the } \\ \text{amount of radium present at the moment. Its half-life is }1590\text{ years. What percentage } \\ \text{will disappear in one year?} \text{[Use }e^{-\frac{\log 2}{1590}}=0.9996\text{].}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original amount of radium be } N \text{ and the amount at time } t \text{ be } P
\displaystyle \text{Given: } \frac{dP}{dt}\propto P
\displaystyle \Rightarrow \frac{dP}{dt}=-aP,\ a>0
\displaystyle \Rightarrow \frac{dP}{P}=-a\,dt
\displaystyle \text{Integrating,}
\displaystyle \int\frac{dP}{P}=\int -a\,dt
\displaystyle \Rightarrow \log|P|=-at+C \qquad (1)
\displaystyle \text{Now, } P=N \text{ at } t=0
\displaystyle \Rightarrow \log|N|=C
\displaystyle \text{Putting } C=\log|N| \text{ in (1),}
\displaystyle \log|P|=-at+\log|N|
\displaystyle \Rightarrow \log\left(\frac{P}{N}\right)=-at \qquad (2)
\displaystyle \text{According to the question, } P=\frac{N}{2} \text{ at } t=1590
\displaystyle \Rightarrow \log\left(\frac{N}{2N}\right)=-1590a
\displaystyle \Rightarrow -\log 2=-1590a
\displaystyle \Rightarrow a=\frac{1}{1590}\log 2
\displaystyle \text{Putting this value of } a \text{ in (2),}
\displaystyle \log\left(\frac{P}{N}\right)=-\frac{\log 2}{1590}t
\displaystyle \Rightarrow \frac{P}{N}=e^{-\frac{\log 2}{1590}t} \qquad (3)
\displaystyle \text{Putting } t=1 \text{ in (3),}
\displaystyle \frac{P}{N}=e^{-\frac{\log 2}{1590}}
\displaystyle \Rightarrow \frac{P}{N}\approx 0.9996
\displaystyle \Rightarrow P\approx 0.9996N
\displaystyle \text{Percentage of amount disappeared in 1 year}=\frac{N-P}{N}\times 100
\displaystyle =0.04\%

\displaystyle \textbf{Solve the following problems based on differential equations:}

\displaystyle \textbf{Question 13: }~\text{The slope of the tangent at a point }P(x,y)\text{ on a curve is }-\frac{x}{y}.\text{ If the curve passes through the point }(3,-4),\text{ find the equation of the curve.}
\displaystyle \text{Answer:}
$\displaystyle \text{According to the question, } \frac{dy}{dx}=-\frac{x}{y}
\displaystyle \Rightarrow y\,dy=-x\,dx
\displaystyle \text{Integrating both sides,}
\displaystyle \int y\,dy=-\int x\,dx
\displaystyle \Rightarrow \frac{y^{2}}{2}=-\frac{x^{2}}{2}+C
\displaystyle \text{Since the curve passes through } (3,-4),
\displaystyle \Rightarrow \frac{(-4)^{2}}{2}=-\frac{3^{2}}{2}+C
\displaystyle \Rightarrow 8=-\frac{9}{2}+C
\displaystyle \Rightarrow C=\frac{25}{2}
\displaystyle \text{Putting the value of } C,
\displaystyle \frac{y^{2}}{2}=-\frac{x^{2}}{2}+\frac{25}{2}
\displaystyle \Rightarrow x^{2}+y^{2}=25

\displaystyle \textbf{Question 14: }~\text{Find the equation of the curve which passes through the point }(2,2) \\ \text{ and satisfies the differential equation }y-x\frac{dy}{dx}=y^{2}+\frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y-x\frac{dy}{dx}=y^{2}+\frac{dy}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}(x+1)=y(1-y)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y(1-y)}{x+1}
\displaystyle \Rightarrow \frac{dy}{y(1-y)}=\frac{dx}{x+1}
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{dy}{y(1-y)}=\int\frac{dx}{x+1}
\displaystyle \Rightarrow \int\left(\frac{1}{y}+\frac{1}{1-y}\right)dy=\int\frac{dx}{x+1}
\displaystyle \Rightarrow \log|y|-\log|1-y|=\log|x+1|+C \qquad (1)
\displaystyle \text{Since the curve passes through } (2,2),
\displaystyle \Rightarrow \log|2|-\log|1-2|=\log|2+1|+C
\displaystyle \Rightarrow \log 2-\log 1=\log 3+C
\displaystyle \Rightarrow C=\log\left(\frac{2}{3}\right)
\displaystyle \text{Putting the value of } C \text{ in (1),}
\displaystyle \log\left|\frac{y}{1-y}\right|=\log\left|\frac{2(x+1)}{3}\right|
\displaystyle \Rightarrow \frac{y}{1-y}=-\frac{2(x+1)}{3}
\displaystyle \Rightarrow 3y=-2(x+1)(1-y)
\displaystyle \Rightarrow 3y=-2x-2+2xy+2y
\displaystyle \Rightarrow 2xy-2x-y-2=0

\displaystyle \textbf{Question 15: }~\text{Find the equation of the curve passing through the point }\left(1,\frac{\pi}{4}\right) \\ \text{ and tangent at any point of which makes an angle }\tan^{-1}\left(\frac{y}{x}-\cos^{2}\frac{y}{x}\right)\text{ with }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The slope of the curve is given as } \frac{dy}{dx}=\tan\theta
\displaystyle \text{Here, } \frac{dy}{dx}=\tan\left(\tan^{-1}\left(\frac{y}{x}\right)-\cos^{2}\left(\frac{y}{x}\right)\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\cos^{2}\left(\frac{y}{x}\right)
\displaystyle \text{Let } y=vx
\displaystyle \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=v-\cos^{2}v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\cos^{2}v
\displaystyle \Rightarrow \sec^{2}v\,dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides,}
\displaystyle \int\sec^{2}v\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \tan v=-\log|x|+C
\displaystyle \Rightarrow \tan\left(\frac{y}{x}\right)=-\log|x|+C
\displaystyle \text{Since the curve passes through } \left(1,\frac{\pi}{4}\right),
\displaystyle \Rightarrow \tan\left(\frac{\pi}{4}\right)=-\log 1+C
\displaystyle \Rightarrow 1=C
\displaystyle \text{Putting the value of } C,
\displaystyle \tan\left(\frac{y}{x}\right)=-\log|x|+1
\displaystyle \Rightarrow \tan\left(\frac{y}{x}\right)=\log\left(\frac{e}{x}\right)

\displaystyle \textbf{Question 16: }~\text{Find the curve for which the intercept cut-off by a tangent on } \\ x\text{-axis is equal to four times the ordinate of the point of contact.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the given curve be } y=f(x). \text{ Suppose } P(x,y) \text{ is a point on the curve.}
\displaystyle \text{Equation of the tangent at } P \text{ is } Y-y=\frac{dy}{dx}(X-x)
\displaystyle \text{Putting } Y=0,
\displaystyle -y=\frac{dy}{dx}(X-x)
\displaystyle \Rightarrow X-x=-y\frac{dx}{dy}
\displaystyle \Rightarrow X=x-y\frac{dx}{dy}
\displaystyle \text{Therefore, intercept on the } x\text{-axis is } x-y\frac{dx}{dy}
\displaystyle \text{Given } x-y\frac{dx}{dy}=4y
\displaystyle \Rightarrow -y\frac{dx}{dy}=4y-x
\displaystyle \Rightarrow \frac{dx}{dy}=\frac{x-4y}{y}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x-4y} \qquad (1)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Put } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx} \text{ in (1)}
\displaystyle v+x\frac{dv}{dx}=\frac{vx}{x-4vx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v}{1-4v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v}{1-4v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{4v^{2}}{1-4v}
\displaystyle \Rightarrow \frac{1-4v}{v^{2}}\,dv=4\frac{dx}{x}
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{1-4v}{v^{2}}\,dv=4\int\frac{dx}{x}
\displaystyle \Rightarrow \int\left(\frac{1}{v^{2}}-\frac{4}{v}\right)dv=4\int\frac{dx}{x}
\displaystyle \Rightarrow -\frac{1}{v}-4\log v=4\log x+\log C
\displaystyle \Rightarrow 4\log(xv)+\log C=-\frac{1}{v}
\displaystyle \text{Putting } v=\frac{y}{x},
\displaystyle 4\log y+\log C=-\frac{x}{y}
\displaystyle \Rightarrow \log(y^{4}C)=-\frac{x}{y}
\displaystyle \Rightarrow y^{4}C=e^{-\frac{x}{y}}

\displaystyle \textbf{Question 17: }~\text{Show that the equation of the curve whose slope at any point is equal to } \\ y+2x\text{ and which passes through the origin is }y+2(x+1)=2e^{2x}.
\displaystyle \text{Answer:}
\displaystyle \text{According to the question, } \frac{dy}{dx}=y+2x
\displaystyle \Rightarrow \frac{dy}{dx}-y=2x \qquad (1)
\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q
\displaystyle \text{where } P=-1 \text{ and } Q=2x
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int(-1)\,dx}
\displaystyle =e^{-x}
\displaystyle \text{Multiplying both sides of (1) by I.F.}=e^{-x}
\displaystyle e^{-x}\left(\frac{dy}{dx}-y\right)=2xe^{-x}
\displaystyle \Rightarrow \frac{d}{dx}(ye^{-x})=2xe^{-x}
\displaystyle \text{Integrating both sides with respect to } x,
\displaystyle ye^{-x}=\int 2xe^{-x}\,dx+C
\displaystyle =2\left(-xe^{-x}-\int(-e^{-x})\,dx\right)+C
\displaystyle =-2xe^{-x}-2e^{-x}+C \qquad (2)
\displaystyle \text{Since the curve passes through the origin,}
\displaystyle \Rightarrow 0=-2\cdot 0-2+C
\displaystyle \Rightarrow C=2
\displaystyle \text{Putting } C=2 \text{ in (2),}
\displaystyle ye^{-x}=-2xe^{-x}-2e^{-x}+2
\displaystyle \Rightarrow y=-2x-2+2e^{x}
\displaystyle \Rightarrow y+2(x+1)=2e^{x}

\displaystyle \textbf{Question 18: }~\text{The tangent at any point }(x,y)\text{ of a curve makes an angle } \\ \tan^{-1}(2x+3y)\text{ with }x\text{-axis. Find the equation of the curve if it passes through }(1,2).
\displaystyle \text{Answer:}
\displaystyle \text{The slope of the curve is given as } \frac{dy}{dx}=\tan\theta
\displaystyle \text{Here, } \theta=\tan^{-1}(2x+3y)
\displaystyle \Rightarrow \frac{dy}{dx}=\tan\left(\tan^{-1}(2x+3y)\right)
\displaystyle \Rightarrow \frac{dy}{dx}=2x+3y
\displaystyle \Rightarrow \frac{dy}{dx}-3y=2x \qquad (1)
\displaystyle \text{This is a linear differential equation of the form } \frac{dy}{dx}+Py=Q
\displaystyle \text{where } P=-3 \text{ and } Q=2x
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int(-3)\,dx}
\displaystyle =e^{-3x}
\displaystyle \text{Multiplying both sides of (1) by I.F.}=e^{-3x}
\displaystyle e^{-3x}\left(\frac{dy}{dx}-3y\right)=2xe^{-3x}
\displaystyle \Rightarrow \frac{d}{dx}(ye^{-3x})=2xe^{-3x}
\displaystyle \text{Integrating both sides with respect to } x,
\displaystyle ye^{-3x}=\int 2xe^{-3x}\,dx+C
\displaystyle =2\left(-\frac{x}{3}e^{-3x}-\int\left(-\frac{1}{3}e^{-3x}\right)dx\right)+C
\displaystyle =-\frac{2x}{3}e^{-3x}-\frac{2}{9}e^{-3x}+C \qquad (2)
\displaystyle \text{Since the curve passes through } (1,2),
\displaystyle \Rightarrow 2e^{-3}=-\frac{2}{3}e^{-3}-\frac{2}{9}e^{-3}+C
\displaystyle \Rightarrow C=\frac{26}{9}e^{-3}
\displaystyle \text{Putting the value of } C \text{ in (2),}
\displaystyle ye^{-3x}=\left(-\frac{2x}{3}-\frac{2}{9}\right)e^{-3x}+\frac{26}{9}e^{-3}

\displaystyle \textbf{Question 19: }~\text{Find the equation of the curve such that the portion of the } x\text{-axis cut off } \\ \text{between the origin and the tangent at a point is twice the abscissa and which passes } \\ \text{through the point }(1,2).
\displaystyle \text{Answer:}
\displaystyle \text{Portion of the } x\text{-axis cut off between the origin and the tangent at a point }=x-y\frac{dx}{dy}
\displaystyle \text{It is given, } x-y\frac{dx}{dy}=2x
\displaystyle \Rightarrow -x=y\frac{dx}{dy}
\displaystyle \Rightarrow -\frac{dx}{x}=\frac{dy}{y}
\displaystyle \text{Integrating,}
\displaystyle -\int\frac{dx}{x}=\int\frac{dy}{y}
\displaystyle \Rightarrow -\log|x|=\log|y|+C
\displaystyle \Rightarrow \log|y|+\log|x|=C
\displaystyle \Rightarrow \log|xy|=C
\displaystyle \Rightarrow xy=k
\displaystyle \text{Since the curve passes through } (1,2),
\displaystyle \Rightarrow 1\cdot 2=k
\displaystyle \Rightarrow k=2
\displaystyle \Rightarrow xy=2

\displaystyle \textbf{Question 20: }~\text{Find the equation to the curve satisfying } \\ x(x+1)\frac{dy}{dx}-y=x(x+1)\text{ and passing through }(1,0).
\displaystyle \text{Answer:}
\displaystyle \text{We have, } x(x+1)\frac{dy}{dx}-y=x(x+1)
\displaystyle \Rightarrow \frac{dy}{dx}-\frac{1}{x(x+1)}y=1
\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q,
\displaystyle P=-\frac{1}{x(x+1)},\quad Q=1
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int-\frac{1}{x(x+1)}\,dx}
\displaystyle =e^{\int\left(-\frac{1}{x}+\frac{1}{x+1}\right)dx}
\displaystyle =e^{-\log|x|+\log|x+1|}
\displaystyle =\frac{x+1}{x}
\displaystyle \text{So, the solution is } y\cdot\text{I.F.}=\int Q\cdot\text{I.F.}\,dx+C
\displaystyle \Rightarrow \frac{x+1}{x}y=\int\frac{x+1}{x}\,dx+C
\displaystyle =\int\left(1+\frac{1}{x}\right)dx+C
\displaystyle =x+\log|x|+C
\displaystyle \text{Since the curve passes through } (1,0),
\displaystyle \Rightarrow 0=1+\log 1+C
\displaystyle \Rightarrow C=-1
\displaystyle \text{Putting } C=-1,
\displaystyle \frac{x+1}{x}y=x+\log|x|-1
\displaystyle \Rightarrow y=\frac{x}{x+1}\left(x+\log|x|-1\right)

\displaystyle \textbf{Question 21: }~\text{Find the equation of the curve which passes through the point }(3,-4) \\ \text{ and has the slope }\frac{2y}{x}\text{ at any point }(x,y)\text{ on it.}
\displaystyle \text{Answer:}
\displaystyle \text{According to the question, } \frac{dy}{dx}=\frac{2y}{x}
\displaystyle \Rightarrow \frac{1}{2y}dy=\frac{1}{x}dx
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{1}{2y}dy=\int\frac{1}{x}dx
\displaystyle \Rightarrow \frac{1}{2}\log|y|=\log|x|+C
\displaystyle \text{Since the curve passes through } (3,-4),
\displaystyle \Rightarrow \frac{1}{2}\log 4=\log 3+C
\displaystyle \Rightarrow C=\log\left(\frac{2}{3}\right)
\displaystyle \text{Putting the value of } C,
\displaystyle \frac{1}{2}\log|y|=\log|x|+\log\left(\frac{2}{3}\right)
\displaystyle \Rightarrow \log|y|=2\log|x|+2\log\left(\frac{2}{3}\right)
\displaystyle \Rightarrow \log|y|=\log\left(\frac{4x^{2}}{9}\right)
\displaystyle \Rightarrow |y|=\frac{4x^{2}}{9}
\displaystyle \Rightarrow y=\pm\frac{4x^{2}}{9}
\displaystyle \text{Given point } (3,-4) \text{ satisfies } y=-\frac{4x^{2}}{9}
\displaystyle \Rightarrow 9y+4x^{2}=0

\displaystyle \textbf{Question 22: }~\text{Find the equation of the curve which passes through the origin and has } \\ \text{the slope } x+3y-1\text{ at any point }(x,y)\text{ on it.}
\displaystyle \text{Answer:}
\displaystyle \text{According to the question, } \frac{dy}{dx}=x+3y-1
\displaystyle \Rightarrow \frac{dy}{dx}-3y=x-1
\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q,
\displaystyle P=-3,\quad Q=x-1
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int(-3)\,dx}=e^{-3x}
\displaystyle \text{So, the solution is } y\cdot\text{I.F.}=\int Q\cdot\text{I.F.}\,dx+C
\displaystyle \Rightarrow ye^{-3x}=\int(x-1)e^{-3x}\,dx+C
\displaystyle =\int xe^{-3x}\,dx-\int e^{-3x}\,dx+C
\displaystyle =\left(-\frac{x}{3}e^{-3x}-\int\left(-\frac{1}{3}e^{-3x}\right)dx\right)+\frac{1}{3}e^{-3x}+C
\displaystyle =-\frac{x}{3}e^{-3x}-\frac{1}{9}e^{-3x}+\frac{1}{3}e^{-3x}+C
\displaystyle =-\frac{x}{3}e^{-3x}+\frac{2}{9}e^{-3x}+C
\displaystyle \Rightarrow y=-\frac{x}{3}+\frac{2}{9}+Ce^{3x}
\displaystyle \text{Since the curve passes through the origin,}
\displaystyle \Rightarrow 0=0+\frac{2}{9}+C
\displaystyle \Rightarrow C=-\frac{2}{9}
\displaystyle \text{Putting } C=-\frac{2}{9},
\displaystyle y=-\frac{x}{3}+\frac{2}{9}\left(1-e^{3x}\right)
\displaystyle \Rightarrow 3(3y+x)=2(1-e^{3x})

\displaystyle \textbf{Question 23: }~\text{At every point on a curve the slope is the sum of the abscissa and the } \\ \text{product  of the ordinate and the abscissa, and the curve passes through }(0,1).\text{ Find } \\ \text{the equation of the curve.}
\displaystyle \text{Answer:}
\displaystyle \text{According to the question, } \frac{dy}{dx}=x+xy
\displaystyle \Rightarrow \frac{dy}{dx}=x(1+y)
\displaystyle \Rightarrow \frac{1}{1+y}dy=x\,dx
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{1}{1+y}dy=\int x\,dx
\displaystyle \Rightarrow \log|1+y|=\frac{x^{2}}{2}+C
\displaystyle \text{Since the curve passes through } (0,1),
\displaystyle \Rightarrow \log 2=0+C
\displaystyle \Rightarrow C=\log 2
\displaystyle \text{Putting the value of } C,
\displaystyle \log|1+y|=\frac{x^{2}}{2}+\log 2
\displaystyle \Rightarrow \log\left(\frac{1+y}{2}\right)=\frac{x^{2}}{2}
\displaystyle \Rightarrow \frac{1+y}{2}=e^{\frac{x^{2}}{2}}
\displaystyle \Rightarrow y+1=2e^{\frac{x^{2}}{2}}

\displaystyle \textbf{Question 24: }~\text{A curve is such that the length of the perpendicular from the origin } \\ \text{on the tangent at any point }P\text{ of the curve is equal to the abscissa of }P.\text{ Prove that the} \\ \text{differential equation of the curve is }y^{2}-2xy\frac{dy}{dx}-x^{2}=0,\text{ and hence find the curve.}
\displaystyle \text{Answer:}
\displaystyle \text{Tangent at } P(x,y) \text{ is given by } Y-y=\frac{dy}{dx}(X-x)
\displaystyle \text{If } p \text{ is the perpendicular from the origin, then}
\displaystyle p=\frac{x\frac{dy}{dx}-y}{\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}}
\displaystyle \text{It is given that } p=x
\displaystyle \Rightarrow \frac{x\frac{dy}{dx}-y}{\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}}=x
\displaystyle \Rightarrow x^{2}\left(\frac{dy}{dx}\right)^{2}-2xy\frac{dy}{dx}+y^{2}=x^{2}+x^{2}\left(\frac{dy}{dx}\right)^{2}
\displaystyle \Rightarrow y^{2}-2xy\frac{dy}{dx}-x^{2}=0
\displaystyle \text{Hence proved.}
\displaystyle \text{Now, } y^{2}-2xy\frac{dy}{dx}-x^{2}=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y^{2}-x^{2}}{2xy}
\displaystyle \Rightarrow 2y\frac{dy}{dx}=\frac{y^{2}}{x}-x
\displaystyle \text{Let } y^{2}=v
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{v}{x}-x
\displaystyle \Rightarrow \frac{dv}{dx}-\frac{v}{x}=-x
\displaystyle \text{I.F.}=e^{\int-\frac{1}{x}dx}=\frac{1}{x}
\displaystyle \text{Multiplying,}
\displaystyle \frac{v}{x}=\int(-x)\frac{1}{x}\,dx+C
\displaystyle =-\int 1\,dx+C
\displaystyle =-x+C
\displaystyle \Rightarrow \frac{y^{2}}{x}=-x+C
\displaystyle \Rightarrow x^{2}+y^{2}=Cx

\displaystyle \textbf{Question 25: }~\text{Find the equation of the curve which passes through the point }(1,2)  \\ \text{ and the distance between the foot of the ordinate of the point of contact and the point } \\ \text{of  intersection of the tangent with }x\text{-axis is twice the abscissa of the point of contact.}
\displaystyle \text{Answer:}
\displaystyle \text{It is given that the distance between the foot of ordinate of the point of contact } A \text{ and the point of intersection of the tangent with the } x\text{-axis } (T) \text{ is } 2x
\displaystyle \text{Coordinate of } T=\left(x-y\frac{dx}{dy},0\right)
\displaystyle AT=\left[x-\left(x-y\frac{dx}{dy}\right)\right]=y\frac{dx}{dy}
\displaystyle \text{Given } AT=2x
\displaystyle \Rightarrow y\frac{dx}{dy}=2x
\displaystyle \Rightarrow \frac{dx}{x}=2\frac{dy}{y}
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{dx}{x}=2\int\frac{dy}{y}
\displaystyle \Rightarrow \log|x|=2\log|y|+\log C
\displaystyle \Rightarrow x=C y^{2}
\displaystyle \text{As the curve passes through } (1,2),
\displaystyle 1=C\cdot 2^{2}
\displaystyle \Rightarrow C=\frac{1}{4}
\displaystyle \Rightarrow 4x=y^{2}

\displaystyle \textbf{Question 26: }~\text{The normal to a given curve at each point }(x,y)\text{ on the curve passes} \\ \text{through the point }(3,0).\text{ If the curve contains the point }(3,4),\text{ find its equation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } P(x,y) \text{ be any point on the curve. The equation of the normal at } P \text{ is}
\displaystyle Y-y=-\frac{1}{\frac{dy}{dx}}(X-x)
\displaystyle \text{It is given that the curve passes through } (3,0)
\displaystyle \Rightarrow 0-y=-\frac{1}{\frac{dy}{dx}}(3-x)
\displaystyle \Rightarrow -y=-\frac{1}{\frac{dy}{dx}}(3-x)
\displaystyle \Rightarrow y\frac{dy}{dx}=3-x
\displaystyle \Rightarrow y\,dy=(3-x)\,dx
\displaystyle \text{Integrating,}
\displaystyle \int y\,dy=\int(3-x)\,dx
\displaystyle \Rightarrow \frac{y^{2}}{2}=3x-\frac{x^{2}}{2}+C \qquad (1)
\displaystyle \text{Since the curve passes through } (3,4),
\displaystyle \Rightarrow \frac{4^{2}}{2}=3(3)-\frac{3^{2}}{2}+C
\displaystyle \Rightarrow 8=9-\frac{9}{2}+C
\displaystyle \Rightarrow C=\frac{7}{2}
\displaystyle \text{Putting } C=\frac{7}{2} \text{ in (1),}
\displaystyle \frac{y^{2}}{2}=3x-\frac{x^{2}}{2}+\frac{7}{2}
\displaystyle \Rightarrow y^{2}=6x-x^{2}+7
\displaystyle \Rightarrow x^{2}+y^{2}-6x-7=0

\displaystyle \textbf{Question 27: }~\text{The rate of increase of bacteria in a culture is proportional to the number } \\ \text{of bacteria present and it is found that the number doubles in }6\text{ hours. Prove that the } \\ \text{bacteria becomes }8\text{ times at the end of }18\text{ hours.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original count of bacteria be } N \text{ and the count at time } t \text{ be } P
\displaystyle \text{Given: } \frac{dP}{dt}\propto P
\displaystyle \Rightarrow \frac{dP}{dt}=aP
\displaystyle \Rightarrow \frac{dP}{P}=a\,dt
\displaystyle \text{Integrating,}
\displaystyle \int\frac{dP}{P}=\int a\,dt
\displaystyle \Rightarrow \log|P|=at+C \qquad (1)
\displaystyle \text{Now, } P=N \text{ at } t=0
\displaystyle \Rightarrow \log|N|=C
\displaystyle \text{Putting } C=\log|N| \text{ in (1),}
\displaystyle \log|P|=at+\log|N|
\displaystyle \Rightarrow \log\left(\frac{P}{N}\right)=at \qquad (2)
\displaystyle \text{According to the question, } \log\left(\frac{2N}{N}\right)=6a
\displaystyle \Rightarrow \log 2=6a
\displaystyle \Rightarrow a=\frac{1}{6}\log 2
\displaystyle \text{Putting this value of } a \text{ in (2),}
\displaystyle \log\left(\frac{P}{N}\right)=\frac{t}{6}\log 2 \qquad (3)
\displaystyle \text{Putting } t=18 \text{ in (3),}
\displaystyle \log\left(\frac{P}{N}\right)=3\log 2
\displaystyle \Rightarrow \log\left(\frac{P}{N}\right)=\log 8
\displaystyle \Rightarrow \frac{P}{N}=8
\displaystyle \Rightarrow P=8N

\displaystyle \textbf{Question 28: }~\text{Radium decomposes at a rate proportional to the quantity of radium} \\ \text{present. It is found that in }25\text{ years, approximately }1.1\%\text{ of a certain quantity of radium } \\ \text{has decomposed. Determine approximately how long it will take for one-half of the } \\ \text{original amount of radium to decompose.}\\ \text{[Given } \log_e 0.989=0.01106\ \text{and }\log_e 2=0.6931\text{].}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original amount of radium be } N \text{ and the amount at time } t \text{ be } P
\displaystyle \text{Given: } \frac{dP}{dt}\propto P
\displaystyle \Rightarrow \frac{dP}{dt}=-aP
\displaystyle \Rightarrow \frac{dP}{P}=-a\,dt
\displaystyle \text{Integrating,}
\displaystyle \int\frac{dP}{P}=\int -a\,dt
\displaystyle \Rightarrow \log|P|=-at+C \qquad (1)
\displaystyle \text{Now, } P=N \text{ at } t=0
\displaystyle \Rightarrow \log|N|=C
\displaystyle \text{Putting } C=\log|N| \text{ in (1),}
\displaystyle \log|P|=-at+\log|N|
\displaystyle \Rightarrow \log\left(\frac{P}{N}\right)=-at \qquad (2)
\displaystyle \text{According to the question, } P=0.989N \text{ at } t=25
\displaystyle \Rightarrow \log\left(\frac{0.989N}{N}\right)=-25a
\displaystyle \Rightarrow \log 0.989=-25a
\displaystyle \Rightarrow a=-\frac{1}{25}\log 0.989
\displaystyle \text{Putting this value of } a \text{ in (2),}
\displaystyle \log\left(\frac{P}{N}\right)=\frac{1}{25}\log 0.989\ t
\displaystyle \text{To find the time when the radium becomes half, take } P=\frac{N}{2}
\displaystyle \Rightarrow \log\left(\frac{N}{2N}\right)=\frac{1}{25}\log 0.989\ t
\displaystyle \Rightarrow -\log 2=\frac{1}{25}\log 0.989\ t
\displaystyle \Rightarrow t=\frac{25\log 2}{-\log 0.989}
\displaystyle \Rightarrow t\approx 1567 \text{ years}

\displaystyle \textbf{Question 29: }~\text{Show that all curves for which the slope at any point }(x,y)\text{ on it is } \\ \frac{x^{2}+y^{2}}{2xy}\text{ are rectangular hyperbola.}
\displaystyle \text{Answer:}
\displaystyle \text{We have, } \frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}
\displaystyle \text{Let } y=vx
\displaystyle \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{x^{2}+v^{2}x^{2}}{2vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1+v^{2}}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{2v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-v^{2}}{2v}
\displaystyle \Rightarrow \frac{2v}{1-v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{2v}{1-v^{2}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\log|1-v^{2}|=\log|x|+C
\displaystyle \Rightarrow \log\left(\frac{1-v^{2}}{C}\right)=-\log|x|
\displaystyle \Rightarrow \frac{1-v^{2}}{C}=\frac{1}{x}
\displaystyle \Rightarrow 1-v^{2}=\frac{C}{x}
\displaystyle \Rightarrow 1-\left(\frac{y}{x}\right)^{2}=\frac{C}{x}
\displaystyle \Rightarrow x^{2}-y^{2}=Cx
\displaystyle \text{Thus, } x^{2}-y^{2}=Cx \text{ is the equation of the rectangular hyperbola.}

\displaystyle \textbf{Question 30: }~\text{The slope of the tangent at each point of a curve is equal to the sum of } \\ \text{the coordinates of the point. Find the curve that passes through the origin.}
\displaystyle \text{Answer:}
\displaystyle \text{According to the question, } \frac{dy}{dx}=x+y
\displaystyle \Rightarrow \frac{dy}{dx}-y=x
\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q,
\displaystyle P=-1,\quad Q=x
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int(-1)\,dx}=e^{-x}
\displaystyle \text{So, the solution is } y\cdot\text{I.F.}=\int Q\cdot\text{I.F.}\,dx+C
\displaystyle \Rightarrow ye^{-x}=\int xe^{-x}\,dx+C
\displaystyle =\left(-xe^{-x}-\int(-e^{-x})\,dx\right)+C
\displaystyle =-xe^{-x}-e^{-x}+C
\displaystyle \text{Since the curve passes through the origin,}
\displaystyle \Rightarrow 0=-0-1+C
\displaystyle \Rightarrow C=1
\displaystyle \text{Putting } C=1,
\displaystyle ye^{-x}=-xe^{-x}-e^{-x}+1
\displaystyle \Rightarrow (y+x+1)e^{-x}=1
\displaystyle \Rightarrow x+y+1=e^{x}

\displaystyle \textbf{Question 31: }~\text{Find the equation of the curve passing through the point }(0,1)\text{ if the } \\ \text{slope of the tangent to the curve at each of its point is equal to the sum of the abscissa} \\ \text{and the product of the abscissa and the ordinate of the point.}
\displaystyle \text{Answer:}
\displaystyle \text{According to the question, } \frac{dy}{dx}=x+xy
\displaystyle \Rightarrow \frac{dy}{dx}-xy=x
\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q, \text{ we get}
\displaystyle P=-x
\displaystyle Q=x
\displaystyle \text{Now,}
\displaystyle \text{I.F.}=e^{\int -x\,dx}
\displaystyle =e^{-\frac{x^{2}}{2}}
\displaystyle \text{So, the solution is given by}
\displaystyle y\times \text{I.F.}=\int Q\times \text{I.F.}\,dx+C
\displaystyle \Rightarrow y e^{-\frac{x^{2}}{2}}=\int x e^{-\frac{x^{2}}{2}}\,dx+C
\displaystyle \Rightarrow y e^{-\frac{x^{2}}{2}}=I+C
\displaystyle \text{Now,}
\displaystyle I=\int x e^{-\frac{x^{2}}{2}}\,dx
\displaystyle \text{Putting } -\frac{x^{2}}{2}=t, \text{ we get}
\displaystyle -x\,dx=dt
\displaystyle \Rightarrow I=-\int e^{t}dt
\displaystyle \Rightarrow I=-e^{t}
\displaystyle \Rightarrow I=-e^{-\frac{x^{2}}{2}}
\displaystyle \Rightarrow y e^{-\frac{x^{2}}{2}}=-e^{-\frac{x^{2}}{2}}+C
\displaystyle \text{Since the curve passes through the point }(0,1), \text{ it satisfies the equation of the curve.}
\displaystyle \Rightarrow 1\cdot e^{0}=-e^{0}+C
\displaystyle \Rightarrow 1=-1+C
\displaystyle \Rightarrow C=2
\displaystyle \text{Putting the value of } C, \text{ we get}
\displaystyle y e^{-\frac{x^{2}}{2}}=-e^{-\frac{x^{2}}{2}}+2
\displaystyle \Rightarrow y=-1+2e^{\frac{x^{2}}{2}}

\displaystyle \textbf{Question 32: }~\text{The slope of a curve at each of its points is equal to the square of the } \\ \text{abscissae of the point. Find the particular curve through the point }(-1,1).
\displaystyle \text{Answer:}
\displaystyle \text{According to the question, } \frac{dy}{dx}=x^{2}
\displaystyle \Rightarrow dy=x^{2}dx
\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}
\displaystyle \int dy=\int x^{2}dx
\displaystyle \Rightarrow y=\frac{x^{3}}{3}+C
\displaystyle \text{Since the curve passes through }(-1,1), \text{ it satisfies the above equation.}
\displaystyle \Rightarrow 1=\frac{(-1)^{3}}{3}+C
\displaystyle \Rightarrow 1=-\frac{1}{3}+C
\displaystyle \Rightarrow C=1+\frac{1}{3}
\displaystyle \Rightarrow C=\frac{4}{3}
\displaystyle \text{Putting the value of } C, \text{ we get}
\displaystyle y=\frac{x^{3}}{3}+\frac{4}{3}
\displaystyle \Rightarrow 3y=x^{3}+4

\displaystyle \textbf{Question 33: }~\text{Find the equation of the curve that passes through the point }(0,a) \\ \text{ and is } \text{such that at any point }(x,y)\text{ on it, the product of its slope and the ordinate is } \\ \text{equal to the abscissa.}
\displaystyle \text{Answer:}
\displaystyle \text{According to the question, } y\frac{dy}{dx}=x
\displaystyle \Rightarrow y\,dy=x\,dx
\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}
\displaystyle \int y\,dy=\int x\,dx
\displaystyle \Rightarrow \frac{y^{2}}{2}=\frac{x^{2}}{2}+C
\displaystyle \text{Since the curve passes through }(0,a), \text{ it satisfies the above equation.}
\displaystyle \Rightarrow \frac{a^{2}}{2}=\frac{0^{2}}{2}+C
\displaystyle \Rightarrow C=\frac{a^{2}}{2}
\displaystyle \text{Putting the value of } C, \text{ we get}
\displaystyle \frac{y^{2}}{2}=\frac{x^{2}}{2}+\frac{a^{2}}{2}
\displaystyle \Rightarrow x^{2}-y^{2}=-a^{2}

\displaystyle \textbf{Question 34: }~\text{The }x\text{-intercept of the tangent line to a curve is equal to the ordinate } \\ \text{of the point of contact. Find the particular curve through the point }(1,1).
\displaystyle \text{Answer:}
\displaystyle \text{Let the curve be }y=y(x)\text{ and let }(x,y)\text{ be any point on it.}
\displaystyle \text{Equation of tangent at }(x,y)\text{ is }Y-y=\frac{dy}{dx}(X-x).
\displaystyle \text{At the }x\text{-intercept of this tangent, }Y=0.
\displaystyle -y=\frac{dy}{dx}(X-x).
\displaystyle X=x-\frac{y}{\frac{dy}{dx}}=x-\frac{y}{y'}. \quad \left(y'=\frac{dy}{dx}\right)
\displaystyle \text{Given: }x\text{-intercept }= \text{ ordinate of the point of contact }=y.
\displaystyle y=x-\frac{y}{y'}.
\displaystyle \frac{y}{y'}=x-y.
\displaystyle y'=\frac{y}{x-y}.
\displaystyle \text{Write }x\text{ as a function of }y:\ \frac{dx}{dy}=\frac{x-y}{y}.
\displaystyle \frac{dx}{dy}-\frac{1}{y}x=-1.
\displaystyle \text{This is linear in }x(y).\ \text{Integrating factor }=\exp\!\left(\int -\frac{1}{y}\,dy\right)=\frac{1}{y}.
\displaystyle \frac{1}{y}\frac{dx}{dy}-\frac{1}{y^{2}}x=-\frac{1}{y}.
\displaystyle \frac{d}{dy}\left(\frac{x}{y}\right)=-\frac{1}{y}.
\displaystyle \frac{x}{y}=-\log|y|+C.
\displaystyle x=y\left(C-\log|y|\right).
\displaystyle \text{Using }(1,1):\ 1=1\left(C-\log 1\right)=C.
\displaystyle \therefore\ C=1.
\displaystyle \text{Hence the required curve is }x=y\left(1-\log y\right).

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