\displaystyle \textbf{Question 1. }\text{Order of the equation }\left(1+5\frac{dy}{dx}\right)^{\frac{3}{2}}=10\frac{d^{3}y}{dx^{3}}\text{ is}
\displaystyle \text{Answer:}
\displaystyle \text{The highest order derivative present is }\frac{d^{3}y}{dx^{3}}
\displaystyle \therefore \text{Order}=3
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 2. }\text{The order and degree of the differential equation } \\ y=x\frac{dy}{dx}+\frac{2}{dy/dx}\text{ are}
\displaystyle \text{Answer:}
\displaystyle y=x\frac{dy}{dx}+\frac{2}{dy/dx}
\displaystyle \therefore y\frac{dy}{dx}=x\left(\frac{dy}{dx}\right)^{2}+2
\displaystyle \therefore x\left(\frac{dy}{dx}\right)^{2}-y\frac{dy}{dx}+2=0
\displaystyle \text{The highest order derivative present is }\frac{dy}{dx}
\displaystyle \therefore \text{Order}=1
\displaystyle \text{The highest power of }\frac{dy}{dx}\text{ is }2
\displaystyle \therefore \text{Degree}=2
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 3. }\text{The degree of the differential equation } \\ x=1+\left(\frac{dy}{dx}\right)+\frac{1}{2!}\left(\frac{dy}{dx}\right)^{2}+\frac{1}{3!}\left(\frac{dy}{dx}\right)^{3}+\ldots\text{ is}
\displaystyle \text{Answer:}
\displaystyle x=1+\left(\frac{dy}{dx}\right)+\frac{1}{2!}\left(\frac{dy}{dx}\right)^{2}+\frac{1}{3!}\left(\frac{dy}{dx}\right)^{3}+\ldots
\displaystyle \therefore x=e^{\frac{dy}{dx}}
\displaystyle \text{Since the differential equation is not a polynomial in derivatives, its degree is not defined.}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 4. }\text{The order of the differential equation whose solution is } \\ y=a\cos x+b\sin x+ce^{-x}\text{ is}
\displaystyle \text{Answer:}
\displaystyle y=a\cos x+b\sin x+ce^{-x}
\displaystyle \text{It contains three arbitrary constants }a,b,c
\displaystyle \therefore \text{Order of the differential equation}=3
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 5. }\text{Solution of }e^{\frac{dy}{dx}}=x,\text{ when }x=1\text{ and }y=0\text{ is}
\displaystyle \text{Answer:}
\displaystyle e^{\frac{dy}{dx}}=x
\displaystyle \therefore \frac{dy}{dx}=\log x
\displaystyle \therefore dy=\log x\,dx
\displaystyle \therefore y=\int\log x\,dx
\displaystyle y=x\log x-x+C
\displaystyle \text{Given }x=1,\ y=0
\displaystyle \therefore 0=1\log 1-1+C
\displaystyle \therefore C=1
\displaystyle \therefore y=x\log x-x+1
\displaystyle =x(\log x-1)+1
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 6. }\text{If }m\text{ and }n\text{ are the order and degree of the differential equation } \\ \left(\frac{d^{2}y}{dx^{2}}\right)^{5}+4\frac{\left(\frac{d^{2}y}{dx^{2}}\right)^{3}}{\frac{d^{3}y}{dx^{3}}}+\frac{d^{3}y}{dx^{3}}=x^{2}-1,\text{ then}
\displaystyle \text{Answer:}
\displaystyle \left(\frac{d^{2}y}{dx^{2}}\right)^{5}+4\frac{\left(\frac{d^{2}y}{dx^{2}}\right)^{3}}{\frac{d^{3}y}{dx^{3}}}+\frac{d^{3}y}{dx^{3}}=x^{2}-1
\displaystyle \text{Multiplying by }\frac{d^{3}y}{dx^{3}},
\displaystyle \left(\frac{d^{2}y}{dx^{2}}\right)^{5}\frac{d^{3}y}{dx^{3}}+4\left(\frac{d^{2}y}{dx^{2}}\right)^{3}+\left(\frac{d^{3}y}{dx^{3}}\right)^{2}=(x^{2}-1)\frac{d^{3}y}{dx^{3}}
\displaystyle \text{The highest order derivative is }\frac{d^{3}y}{dx^{3}}
\displaystyle \therefore m=3
\displaystyle \text{The highest power of }\frac{d^{3}y}{dx^{3}}\text{ is }2
\displaystyle \therefore n=2
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 7. }\text{The order of the differential equation satisfying } \\ \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)\text{ is}
\displaystyle \text{Answer:}
\displaystyle \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)
\displaystyle \text{It contains one arbitrary constant }a
\displaystyle \therefore \text{Order of the differential equation}=1
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 8. }\text{The solution of }\frac{dy}{dx}-y=1,\ y(0)=1\text{ is given by}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}-y=1
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=-1,\ Q=1
\displaystyle \text{Integrating factor }=e^{\int -1\,dx}=e^{-x}
\displaystyle \therefore ye^{-x}=\int e^{-x}\,dx+C
\displaystyle ye^{-x}=-e^{-x}+C
\displaystyle \therefore y=-1+Ce^{x}
\displaystyle \text{Given }y(0)=1
\displaystyle \therefore 1=-1+C
\displaystyle \therefore C=2
\displaystyle \therefore y=2e^{x}-1
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 9. }\text{The differential equation }y\frac{dy}{dx}+x=C\text{ represents}
\displaystyle \text{Answer:}
\displaystyle y\frac{dy}{dx}+x=C
\displaystyle \therefore y\,dy=(C-x)\,dx
\displaystyle \therefore \int y\,dy=\int(C-x)\,dx
\displaystyle \frac{y^{2}}{2}=Cx-\frac{x^{2}}{2}+K
\displaystyle \therefore x^{2}+y^{2}-2Cx=K_{1}
\displaystyle \therefore (x-C)^{2}+y^{2}=K_{2}
\displaystyle \text{This is the equation of a circle.}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 10. }\text{The integrating factor of differential equation } \\ \frac{dy}{dx}+y\tan x-\sec x=0\text{ is}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}+y\tan x=\sec x
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=\tan x
\displaystyle \text{Integrating factor}=e^{\int P\,dx}
\displaystyle =e^{\int\tan x\,dx}
\displaystyle =e^{\log(\sec x)}
\displaystyle =\sec x
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 11. }\text{State whether }y=e^{-x}(x+a)\text{ is a solution of differential equation } \\ \frac{dy}{dx}+y=e^{-x}
\displaystyle \text{Answer:}
\displaystyle y=e^{-x}(x+a)
\displaystyle \therefore \frac{dy}{dx}=e^{-x}\cdot 1+(x+a)(-e^{-x})
\displaystyle =e^{-x}-(x+a)e^{-x}
\displaystyle \frac{dy}{dx}+y=e^{-x}-(x+a)e^{-x}+e^{-x}(x+a)
\displaystyle =e^{-x}
\displaystyle \therefore y=e^{-x}(x+a)\text{ is a solution of the given differential equation.}
\\

\displaystyle \textbf{Question 12. }\text{Write the solution of differential equation } \\ (e^{x}+e^{-x})dy=(e^{x}-e^{-x})dx
\displaystyle \text{Answer:}
\displaystyle (e^{x}+e^{-x})dy=(e^{x}-e^{-x})dx
\displaystyle \therefore dy=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx
\displaystyle \therefore y=\int\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx
\displaystyle \text{Let }t=e^{x}+e^{-x}
\displaystyle \therefore dt=(e^{x}-e^{-x})dx
\displaystyle \therefore y=\int\frac{dt}{t}
\displaystyle y=\log|t|+C
\displaystyle \therefore y=\log|e^{x}+e^{-x}|+C
\\

\displaystyle \textbf{Question 13. }\text{Solve the differential equation } \\ \frac{dy}{dx}=y\sin 2x\text{ given that }y(0)=1
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=y\sin 2x
\displaystyle \therefore \frac{dy}{y}=\sin 2x\,dx
\displaystyle \therefore \int\frac{dy}{y}=\int\sin 2x\,dx
\displaystyle \log y=-\frac{\cos 2x}{2}+C
\displaystyle \text{Given }y(0)=1
\displaystyle \therefore \log 1=-\frac{\cos 0}{2}+C
\displaystyle 0=-\frac{1}{2}+C
\displaystyle \therefore C=\frac{1}{2}
\displaystyle \therefore \log y=\frac{1-\cos 2x}{2}
\displaystyle =\sin^{2}x
\displaystyle \therefore y=e^{\sin^{2}x}
\\

\displaystyle \textbf{Question 14. }\text{Solve }2(y+3)-xy\frac{dy}{dx}=0,\text{ given that }y(1)=-2
\displaystyle \text{Answer:}
\displaystyle 2(y+3)-xy\frac{dy}{dx}=0
\displaystyle \therefore xy\frac{dy}{dx}=2(y+3)
\displaystyle \therefore \frac{y}{y+3}\,dy=\frac{2}{x}\,dx
\displaystyle \therefore \int\frac{y}{y+3}\,dy=\int\frac{2}{x}\,dx
\displaystyle \int\left(1-\frac{3}{y+3}\right)dy=2\log|x|+C
\displaystyle y-3\log|y+3|=2\log|x|+C
\displaystyle \text{Given }y(1)=-2
\displaystyle \therefore -2-3\log 1=2\log 1+C
\displaystyle \therefore C=-2
\displaystyle \therefore y-3\log|y+3|=2\log|x|-2
\\

\displaystyle \textbf{Question 15. }\text{Find the integrating factor of the differential equation } \\ e^{2x}\frac{dy}{dx}+3e^{2x}y=1
\displaystyle \text{Answer:}
\displaystyle e^{2x}\frac{dy}{dx}+3e^{2x}y=1
\displaystyle \therefore \frac{dy}{dx}+3y=e^{-2x}
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=3
\displaystyle \text{Integrating factor }=e^{\int P\,dx}
\displaystyle =e^{\int 3\,dx}
\displaystyle =e^{3x}
\\

\displaystyle \textbf{Question 16. }\text{State whether }y=e^{-x}(x+a)\text{ is a solution of differential equation } \\ \frac{dy}{dx}+y=e^{-x}
\displaystyle \text{Answer:}
\displaystyle y=e^{-x}(x+a)
\displaystyle \therefore \frac{dy}{dx}=e^{-x}-(x+a)e^{-x}
\displaystyle \frac{dy}{dx}+y=e^{-x}-(x+a)e^{-x}+e^{-x}(x+a)
\displaystyle =e^{-x}
\displaystyle \therefore y=e^{-x}(x+a)\text{ is a solution of the given differential equation.}
\\

\displaystyle \textbf{Question 17. }\text{Show that the function }y=(A+Bx)e^{3x}\text{ is a solution of the equation } \\ \frac{d^{2}y}{dx^{2}}-6\frac{dy}{dx}+9y=0
\displaystyle \text{Answer:}
\displaystyle y=(A+Bx)e^{3x}
\displaystyle \therefore \frac{dy}{dx}=Be^{3x}+3(A+Bx)e^{3x}
\displaystyle =(3A+B+3Bx)e^{3x}
\displaystyle \therefore \frac{d^{2}y}{dx^{2}}=3Be^{3x}+3(3A+B+3Bx)e^{3x}
\displaystyle =(9A+6B+9Bx)e^{3x}
\displaystyle \therefore \frac{d^{2}y}{dx^{2}}-6\frac{dy}{dx}+9y
\displaystyle =(9A+6B+9Bx)e^{3x}-6(3A+B+3Bx)e^{3x}+9(A+Bx)e^{3x}
\displaystyle =e^{3x}\left[9A+6B+9Bx-18A-6B-18Bx+9A+9Bx\right]
\displaystyle =0
\displaystyle \therefore y=(A+Bx)e^{3x}\text{ is a solution of the given differential equation.}
\\

\displaystyle \textbf{Question 18. }\text{Solve }(x-1)\frac{dy}{dx}=2x^{3}y
\displaystyle \text{Answer:}
\displaystyle (x-1)\frac{dy}{dx}=2x^{3}y
\displaystyle \therefore \frac{dy}{y}=\frac{2x^{3}}{x-1}\,dx
\displaystyle \frac{2x^{3}}{x-1}=2x^{2}+2x+2+\frac{2}{x-1}
\displaystyle \therefore \int\frac{dy}{y}=\int\left(2x^{2}+2x+2+\frac{2}{x-1}\right)dx
\displaystyle \log|y|=\frac{2x^{3}}{3}+x^{2}+2x+2\log|x-1|+C
\\

\displaystyle \textbf{Question 19. }\text{Find the particular solution of the differential equation } \\ x(1+y^{2})dx-y(1+x^{2})dy=0,\text{ given that }y=1,\text{ when }x=0
\displaystyle \text{Answer:}
\displaystyle x(1+y^{2})dx-y(1+x^{2})dy=0
\displaystyle \therefore x(1+y^{2})dx=y(1+x^{2})dy
\displaystyle \therefore \frac{x}{1+x^{2}}\,dx=\frac{y}{1+y^{2}}\,dy
\displaystyle \therefore \int\frac{x}{1+x^{2}}\,dx=\int\frac{y}{1+y^{2}}\,dy
\displaystyle \frac{1}{2}\log(1+x^{2})=\frac{1}{2}\log(1+y^{2})+C
\displaystyle \therefore \log(1+x^{2})-\log(1+y^{2})=C
\displaystyle \text{Given }x=0,\ y=1
\displaystyle \therefore \log 1-\log 2=C
\displaystyle \therefore C=-\log 2
\displaystyle \therefore \log\left(\frac{1+x^{2}}{1+y^{2}}\right)=-\log 2
\displaystyle \therefore \frac{1+x^{2}}{1+y^{2}}=\frac{1}{2}
\displaystyle \therefore 1+y^{2}=2(1+x^{2})
\displaystyle \therefore y^{2}=2x^{2}+1
\\

\displaystyle \textbf{Question 20. }\text{Find the general solution of the differential equation } \\ e^{2x}\frac{dy}{dx}+3e^{2x}y=1
\displaystyle \text{Answer:}
\displaystyle e^{2x}\frac{dy}{dx}+3e^{2x}y=1
\displaystyle \therefore \frac{dy}{dx}+3y=e^{-2x}
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=3,\ Q=e^{-2x}
\displaystyle \text{Integrating factor }=e^{\int P\,dx}=e^{3x}
\displaystyle \therefore y\cdot e^{3x}=\int e^{3x}\cdot e^{-2x}\,dx+C
\displaystyle =\int e^{x}\,dx+C
\displaystyle =e^{x}+C
\displaystyle \therefore y=e^{-2x}+Ce^{-3x}
\\

\displaystyle \textbf{Question 21. }\text{Solve the differential equation }x\,dy=(2y+2x^{4}+x^{2})\,dx
\displaystyle \text{Answer:}
\displaystyle x\frac{dy}{dx}=2y+2x^{4}+x^{2}
\displaystyle \therefore \frac{dy}{dx}-\frac{2}{x}y=2x^{3}+x
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=-\frac{2}{x},\ Q=2x^{3}+x
\displaystyle \text{Integrating factor }=e^{\int P\,dx}=e^{\int-\frac{2}{x}\,dx}
\displaystyle =e^{-2\log x}=x^{-2}
\displaystyle \therefore yx^{-2}=\int x^{-2}(2x^{3}+x)\,dx+C
\displaystyle =\int\left(2x+\frac{1}{x}\right)dx+C
\displaystyle =x^{2}+\log|x|+C
\displaystyle \therefore y=x^{2}(x^{2}+\log|x|+C)
\\

\displaystyle \textbf{Question 22. }\text{Solve the differential equation } \\ \cos x\cdot\frac{dy}{dx}+2\sin x\cdot y=\sin x\cdot\cos x
\displaystyle \text{Answer:}
\displaystyle \cos x\frac{dy}{dx}+2\sin x\,y=\sin x\cos x
\displaystyle \therefore \frac{dy}{dx}+2\tan x\,y=\sin x
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=2\tan x,\ Q=\sin x
\displaystyle \text{Integrating factor }=e^{\int 2\tan x\,dx}
\displaystyle =e^{2\log|\sec x|}=\sec^{2}x
\displaystyle \therefore y\sec^{2}x=\int\sec^{2}x\sin x\,dx+C
\displaystyle =\int\frac{\sin x}{\cos^{2}x}\,dx+C
\displaystyle =\sec x+C
\displaystyle \therefore y=(\sec x+C)\cos^{2}x
\displaystyle \therefore y=\cos x+C\cos^{2}x
\\

\displaystyle \textbf{Question 23. }\text{Solve the differential equation }\frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}
\displaystyle \therefore e^{y}\frac{dy}{dx}=e^{x}+x^{2}
\displaystyle \therefore \frac{d}{dx}(e^{y})=e^{x}+x^{2}
\displaystyle \therefore e^{y}=\int(e^{x}+x^{2})\,dx
\displaystyle \therefore e^{y}=e^{x}+\frac{x^{3}}{3}+C
\\

\displaystyle \textbf{Question 24. }\text{Solve the differential equation }y+x\frac{dy}{dx}=x-y\frac{dy}{dx}
\displaystyle \text{Answer:}
\displaystyle y+x\frac{dy}{dx}=x-y\frac{dy}{dx}
\displaystyle \therefore (x+y)\frac{dy}{dx}=x-y
\displaystyle \therefore \frac{dy}{dx}=\frac{x-y}{x+y}
\displaystyle \text{This is a homogeneous differential equation. Let }y=vx
\displaystyle \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=\frac{1-v}{1+v}
\displaystyle x\frac{dv}{dx}=\frac{1-v-v-v^{2}}{1+v}
\displaystyle x\frac{dv}{dx}=\frac{1-2v-v^{2}}{1+v}
\displaystyle \therefore \frac{1+v}{1-2v-v^{2}}\,dv=\frac{dx}{x}
\displaystyle \text{Let }1-2v-v^{2}=t
\displaystyle \therefore -(2+2v)dv=dt
\displaystyle \therefore (1+v)dv=-\frac{1}{2}dt
\displaystyle -\frac{1}{2}\int\frac{dt}{t}=\int\frac{dx}{x}
\displaystyle -\frac{1}{2}\log|t|=\log|x|+C
\displaystyle \log|1-2v-v^{2}|=-2\log|x|+C
\displaystyle x^{2}(1-2v-v^{2})=C
\displaystyle \text{Putting }v=\frac{y}{x},
\displaystyle x^{2}\left(1-\frac{2y}{x}-\frac{y^{2}}{x^{2}}\right)=C
\displaystyle \therefore x^{2}-2xy-y^{2}=C
\\

\displaystyle \textbf{Question 25. }\text{Solve the differential equation }(y+3x^{2})\frac{dx}{dy}=x
\displaystyle \text{Answer:}
\displaystyle (y+3x^{2})\frac{dx}{dy}=x
\displaystyle \therefore \frac{dy}{dx}=\frac{y+3x^{2}}{x}
\displaystyle \therefore \frac{dy}{dx}-\frac{1}{x}y=3x
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=-\frac{1}{x},\ Q=3x
\displaystyle \text{Integrating factor }=e^{\int-\frac{1}{x}\,dx}
\displaystyle =e^{-\log x}=\frac{1}{x}
\displaystyle \therefore \frac{y}{x}=\int\frac{1}{x}\cdot3x\,dx+C
\displaystyle =3x+C
\displaystyle \therefore y=3x^{2}+Cx
\\

\displaystyle \textbf{Question 26. }\text{Solve the differential equation }dy=\cos x(2-y\,\mathrm{cosec}\,x)dx \\ \text{ given that }y=2,\text{ when }x=\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=\cos x(2-y\,\mathrm{cosec}\,x)
\displaystyle \therefore \frac{dy}{dx}=2\cos x-y\cot x
\displaystyle \therefore \frac{dy}{dx}+y\cot x=2\cos x
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=\cot x,\ Q=2\cos x
\displaystyle \text{Integrating factor }=e^{\int\cot x\,dx}
\displaystyle =e^{\log|\sin x|}=\sin x
\displaystyle \therefore y\sin x=\int 2\sin x\cos x\,dx+C
\displaystyle y\sin x=\sin^{2}x+C
\displaystyle \text{Given }y=2\text{ when }x=\frac{\pi}{2}
\displaystyle \therefore 2\cdot1=1+C
\displaystyle \therefore C=1
\displaystyle \therefore y\sin x=\sin^{2}x+1
\\

\displaystyle \textbf{Question 27. }\text{Show that the differential equation }(y^{2}-x^{2})dy=3xy\,dx \\ \text{is homogeneous and solve it}
\displaystyle \text{Answer:}
\displaystyle (y^{2}-x^{2})dy=3xy\,dx
\displaystyle \therefore \frac{dy}{dx}=\frac{3xy}{y^{2}-x^{2}}
\displaystyle =\frac{3\left(\frac{y}{x}\right)}{\left(\frac{y}{x}\right)^{2}-1}
\displaystyle \therefore \text{The differential equation is homogeneous.}
\displaystyle \text{Let }y=vx
\displaystyle \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=\frac{3v}{v^{2}-1}
\displaystyle x\frac{dv}{dx}=\frac{3v}{v^{2}-1}-v
\displaystyle =\frac{4v-v^{3}}{v^{2}-1}
\displaystyle \therefore \frac{v^{2}-1}{v(4-v^{2})}\,dv=\frac{dx}{x}
\displaystyle \int\frac{v^{2}-1}{v(4-v^{2})}\,dv=\int\frac{dx}{x}
\displaystyle \int\left[-\frac{1}{4v}+\frac{3v}{4(4-v^{2})}\right]dv=\log|x|+C
\displaystyle -\frac{1}{4}\log|v|-\frac{3}{8}\log|4-v^{2}|=\log|x|+C
\displaystyle 2\log|v|+3\log|4-v^{2}|=-8\log|x|+C
\displaystyle v^{2}(4-v^{2})^{3}x^{8}=C
\displaystyle \text{Putting }v=\frac{y}{x}
\displaystyle \left(\frac{y^{2}}{x^{2}}\right)\left(4-\frac{y^{2}}{x^{2}}\right)^{3}x^{8}=C
\displaystyle \therefore y^{2}(4x^{2}-y^{2})^{3}=C
\\

\displaystyle \textbf{Question 28. }\text{Solve the equation }x\,dy-y\,dx=\sqrt{x^{2}+y^{2}}\,dx
\displaystyle \text{Answer:}
\displaystyle x\frac{dy}{dx}-y=\sqrt{x^{2}+y^{2}}
\displaystyle \therefore x\frac{dy}{dx}=y+\sqrt{x^{2}+y^{2}}
\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^{2}}
\displaystyle \text{This is a homogeneous differential equation. Let }y=vx
\displaystyle \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=v+\sqrt{1+v^{2}}
\displaystyle \therefore x\frac{dv}{dx}=\sqrt{1+v^{2}}
\displaystyle \therefore \frac{dv}{\sqrt{1+v^{2}}}=\frac{dx}{x}
\displaystyle \therefore \log|v+\sqrt{1+v^{2}}|=\log|x|+C
\displaystyle \therefore v+\sqrt{1+v^{2}}=Cx
\displaystyle \text{Putting }v=\frac{y}{x},
\displaystyle \frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}=Cx
\displaystyle \therefore y+\sqrt{x^{2}+y^{2}}=Cx^{2}
\\

\displaystyle \textbf{Question 29. }\text{Find the particular solution of the differential equation } \\ x(x^{2}-1)\frac{dy}{dx}=1,\ y=0\text{ when }x=2
\displaystyle \text{Answer:}
\displaystyle x(x^{2}-1)\frac{dy}{dx}=1
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{x(x^{2}-1)}
\displaystyle =\frac{1}{x(x-1)(x+1)}
\displaystyle \frac{1}{x(x-1)(x+1)}=-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}
\displaystyle \therefore y=\int\left[-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\right]dx
\displaystyle y=-\log|x|+\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|+C
\displaystyle y=\frac{1}{2}\log|x^{2}-1|-\log|x|+C
\displaystyle \text{Given }y=0\text{ when }x=2
\displaystyle \therefore 0=\frac{1}{2}\log 3-\log 2+C
\displaystyle \therefore C=\log 2-\frac{1}{2}\log 3
\displaystyle \therefore y=\frac{1}{2}\log|x^{2}-1|-\log|x|+\log 2-\frac{1}{2}\log 3
\\

\displaystyle \textbf{Question 30. }\text{Solve the differential equation }x\frac{dy}{dx}=y-x\tan\left(\frac{y}{x}\right)
\displaystyle \text{Answer:}
\displaystyle x\frac{dy}{dx}=y-x\tan\left(\frac{y}{x}\right)
\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}-\tan\left(\frac{y}{x}\right)
\displaystyle \text{This is a homogeneous differential equation. Let }y=vx
\displaystyle \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=v-\tan v
\displaystyle \therefore x\frac{dv}{dx}=-\tan v
\displaystyle \therefore \cot v\,dv=-\frac{dx}{x}
\displaystyle \therefore \int\cot v\,dv=-\int\frac{dx}{x}
\displaystyle \log|\sin v|=-\log|x|+C
\displaystyle \therefore x\sin v=C
\displaystyle \text{Putting }v=\frac{y}{x},
\displaystyle x\sin\left(\frac{y}{x}\right)=C
\\

\displaystyle \textbf{Question 31. }\text{Solve the differential equation }(1+x^{2})\frac{dy}{dx}+y=\tan^{-1}x
\displaystyle \text{Answer:}
\displaystyle (1+x^{2})\frac{dy}{dx}+y=\tan^{-1}x
\displaystyle \therefore \frac{dy}{dx}+\frac{1}{1+x^{2}}y=\frac{\tan^{-1}x}{1+x^{2}}
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=\frac{1}{1+x^{2}},\ Q=\frac{\tan^{-1}x}{1+x^{2}}
\displaystyle \text{Integrating factor }=e^{\int\frac{1}{1+x^{2}}\,dx}
\displaystyle =e^{\tan^{-1}x}
\displaystyle \therefore ye^{\tan^{-1}x}=\int e^{\tan^{-1}x}\frac{\tan^{-1}x}{1+x^{2}}\,dx+C
\displaystyle \text{Let }t=\tan^{-1}x
\displaystyle \therefore dt=\frac{1}{1+x^{2}}\,dx
\displaystyle ye^{\tan^{-1}x}=\int te^{t}\,dt+C
\displaystyle =e^{t}(t-1)+C
\displaystyle =e^{\tan^{-1}x}(\tan^{-1}x-1)+C
\displaystyle \therefore y=\tan^{-1}x-1+Ce^{-\tan^{-1}x}
\\

\displaystyle \textbf{Question 32. }\text{Solve }\frac{dy}{dx}=\cos(x+y)+\sin(x+y)
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=\cos(x+y)+\sin(x+y)
\displaystyle \text{Let }u=x+y
\displaystyle \therefore \frac{du}{dx}=1+\frac{dy}{dx}
\displaystyle =1+\cos u+\sin u
\displaystyle \therefore \frac{du}{1+\sin u+\cos u}=dx
\displaystyle \text{Let }t=\tan\frac{u}{2}
\displaystyle \therefore \sin u=\frac{2t}{1+t^{2}},\ \cos u=\frac{1-t^{2}}{1+t^{2}},\ du=\frac{2}{1+t^{2}}dt
\displaystyle \therefore \int\frac{du}{1+\sin u+\cos u}=\int\frac{dt}{1+t}
\displaystyle \therefore \log|1+t|=x+C
\displaystyle \therefore \log\left|1+\tan\frac{u}{2}\right|=x+C
\displaystyle \therefore \log\left|1+\tan\frac{x+y}{2}\right|=x+C
\\

\displaystyle \textbf{Question 33. }\text{Solve the differential equation }\frac{dy}{dx}-3y\cot x=\sin 2x, \\ \text{ given }y=2,\text{ when }x=\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}-3y\cot x=\sin 2x
\displaystyle \text{This is of the form }\frac{dy}{dx}+Py=Q
\displaystyle \therefore P=-3\cot x,\ Q=\sin 2x
\displaystyle \text{Integrating factor }=e^{\int -3\cot x\,dx}
\displaystyle =e^{-3\log|\sin x|}=\mathrm{cosec}^{3}x
\displaystyle \therefore y\,\mathrm{cosec}^{3}x=\int\sin 2x\,\mathrm{cosec}^{3}x\,dx+C
\displaystyle =\int 2\sin x\cos x\,\mathrm{cosec}^{3}x\,dx+C
\displaystyle =\int 2\cos x\,\mathrm{cosec}^{2}x\,dx+C
\displaystyle =-2\mathrm{cosec}\,x+C
\displaystyle \text{Given }y=2\text{ when }x=\frac{\pi}{2}
\displaystyle \therefore 2=-2+C
\displaystyle \therefore C=4
\displaystyle \therefore y\,\mathrm{cosec}^{3}x=-2\mathrm{cosec}\,x+4
\displaystyle \therefore y=-2\sin^{2}x+4\sin^{3}x
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\displaystyle \textbf{Question 34. }\text{Solve the differential equation } \\ \left[y-x\cos\left(\frac{y}{x}\right)\right]dy+\left[y\cos\left(\frac{y}{x}\right)-2x\sin\left(\frac{y}{x}\right)\right]dx=0
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=vx
\displaystyle \therefore dy=v\,dx+x\,dv
\displaystyle \left[xv-x\cos v\right](v\,dx+x\,dv)+\left[xv\cos v-2x\sin v\right]dx=0
\displaystyle x(v-\cos v)(v\,dx+x\,dv)+x(v\cos v-2\sin v)dx=0
\displaystyle x(v^{2}-2\sin v)dx+x^{2}(v-\cos v)dv=0
\displaystyle \therefore \frac{dx}{x}=\frac{v-\cos v}{2\sin v-v^{2}}\,dv
\displaystyle \text{Let }t=2\sin v-v^{2}
\displaystyle \therefore dt=2(\cos v-v)dv=-2(v-\cos v)dv
\displaystyle \therefore \int\frac{dx}{x}=-\frac{1}{2}\int\frac{dt}{t}
\displaystyle \log|x|=-\frac{1}{2}\log|t|+C
\displaystyle \therefore x^{2}t=C
\displaystyle \therefore x^{2}\left(2\sin v-v^{2}\right)=C
\displaystyle \text{Putting }v=\frac{y}{x},
\displaystyle x^{2}\left(2\sin\left(\frac{y}{x}\right)-\frac{y^{2}}{x^{2}}\right)=C
\displaystyle \therefore 2x^{2}\sin\left(\frac{y}{x}\right)-y^{2}=C
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