\displaystyle \textbf{Form the differential equations in the following questions:}

\displaystyle \textbf{Question 1: }~\text{Form the differential equation of the family of curves represented } \\ \text{by }y^{2}=(x-c)^{3}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y^2=(x-c)^3\quad\text{(1)}
\displaystyle \text{where }c\in R\text{ is a parameter.}
\displaystyle \text{This equation contains only one parameter, so we shall obtain a differential} \\ \text{equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2y\frac{dy}{dx}=3(x-c)^2\quad\text{(2)}
\displaystyle \Rightarrow \frac{y^2}{2y\frac{dy}{dx}}=\frac{(x-c)^3}{3(x-c)^2}
\displaystyle \Rightarrow \frac{y}{2\frac{dy}{dx}}=\frac{x-c}{3}
\displaystyle \Rightarrow \frac{3y}{2\frac{dy}{dx}}=x-c
\displaystyle \Rightarrow c=x-\frac{3y}{2\frac{dy}{dx}}
\displaystyle \text{Substituting the value of }c\text{ in equation (1), we get}
\displaystyle y^2=\left(x-x+\frac{3y}{2\frac{dy}{dx}}\right)^3
\displaystyle \Rightarrow y^2=\frac{27y^3}{8\left(\frac{dy}{dx}\right)^3}
\displaystyle \Rightarrow 8y^2\left(\frac{dy}{dx}\right)^3=27y^3
\displaystyle \Rightarrow 8\left(\frac{dy}{dx}\right)^3-27y=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 2: }~\text{Form the differential equation corresponding to }y=e^{mx}\text{ by } \\ \text{eliminating }m.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y=e^{mx}\quad\text{(1)}
\displaystyle \text{where }m\text{ is a parameter.}
\displaystyle \text{This equation contains only one parameter, so we shall get a differential } \\ \text{equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{dy}{dx}=me^{mx}
\displaystyle \Rightarrow \frac{dy}{dx}=my\quad\text{[Using equation (1)]}
\displaystyle \Rightarrow m=\frac{1}{y}\frac{dy}{dx}\quad\text{(2)}
\displaystyle \text{Now, from equation (1), we get}
\displaystyle \log y=\log e^{mx}
\displaystyle \Rightarrow \log y=mx\log e
\displaystyle \Rightarrow \log y=mx
\displaystyle \Rightarrow m=\frac{1}{x}\log y\quad\text{(3)}
\displaystyle \text{Comparing equations (2) and (3), we get}
\displaystyle \frac{1}{x}\log y=\frac{1}{y}\frac{dy}{dx}
\displaystyle \Rightarrow x\frac{dy}{dx}=y\log y
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 3: }~\text{Form the differential equations from the following primitives } \\ \text{where constants are arbitrary:}
\displaystyle \text{(i): }y^{2}=4ax. \hspace{5.0cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y^2=4ax\quad\text{(1)}
\displaystyle \text{where }a\text{ is an arbitrary constant.}
\displaystyle \text{This equation contains only one arbitrary constant, so we shall get a differential } \\ \text{equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2y\frac{dy}{dx}=4a
\displaystyle \Rightarrow a=\frac{y}{2}\frac{dy}{dx}\quad\text{(2)}
\displaystyle \text{Putting the value of }a\text{ from equation (2) in equation (1), we get}
\displaystyle y^2=4x\left(\frac{y}{2}\frac{dy}{dx}\right)
\displaystyle \Rightarrow y=2x\frac{dy}{dx}
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(ii): }y=cx+2c^{2}+c^{3}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y=cx+2c^2+c^3\quad\text{(1)}
\displaystyle \text{where }c\text{ is an arbitrary constant.}
\displaystyle \text{This equation contains only one arbitrary constant, so we shall get a differential } \\ \text{equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{dy}{dx}=c\quad\text{(2)}
\displaystyle \text{Putting the value of }c\text{ from equation (2) in equation (1), we get}
\displaystyle y=x\frac{dy}{dx}+2\left(\frac{dy}{dx}\right)^2+\left(\frac{dy}{dx}\right)^3
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(iii): }xy=a^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }xy=a^2\quad\text{(1)}
\displaystyle \text{where }a\text{ is an arbitrary constant.}
\displaystyle \text{This equation contains only one arbitrary constant, so we shall get a differential } \\ \text{equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle y+x\frac{dy}{dx}=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(iv): }y=ax^{2}+bx+c.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y=ax^2+bx+c\quad\text{(1)}
\displaystyle \text{where }a,\ b\text{ and }c\text{ are arbitrary constants. So, we shall get a differential } \\ \text{equation of third order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{dy}{dx}=2ax+b\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=2a\quad\text{(3)}
\displaystyle \text{Differentiating equation (3) with respect to }x\text{, we get}
\displaystyle \frac{d^3y}{dx^3}=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 4: }~\text{Find the differential equation of the family of curves, } \\ y=Ae^{2x}+Be^{-2x},\text{ where }A\text{ and }B\text{ are arbitrary constants.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y=Ae^{2x}+Be^{-2x}\quad\text{(1)}
\displaystyle \text{where }A\text{ and }B\text{ are arbitrary constants.}
\displaystyle \text{This equation contains two arbitrary constants, so we shall get a differential } \\ \text{equation of second order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{dy}{dx}=2Ae^{2x}-2Be^{-2x}\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=4Ae^{2x}+4Be^{-2x}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=4\left(Ae^{2x}+Be^{-2x}\right)
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=4y
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 5: }~\text{Find the differential equation of the family of curves, } \\ x=A\cos nt+B\sin nt,\text{ where }A\text{ and }B\text{ are arbitrary constants.} \hspace{2.0cm} \text{[CBSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }x=A\cos nt+B\sin nt\quad\text{(1)}
\displaystyle \text{where }A\text{ and }B\text{ are arbitrary constants.}
\displaystyle \text{This equation contains two arbitrary constants, so we shall get a differential } \\ \text{equation of second order.}
\displaystyle \text{Differentiating equation (1) with respect to }t\text{, we get}
\displaystyle \frac{dx}{dt}=-An\sin nt+Bn\cos nt\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }t\text{, we get}
\displaystyle \frac{d^2x}{dt^2}=-An^2\cos nt-Bn^2\sin nt
\displaystyle \Rightarrow \frac{d^2x}{dt^2}=-n^2(A\cos nt+B\sin nt)
\displaystyle \Rightarrow \frac{d^2x}{dt^2}=-n^2x
\displaystyle \Rightarrow \frac{d^2x}{dt^2}+n^2x=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 6: }~\text{Form the differential equation corresponding to } \\ y^{2}=a(b-x^{2})\text{ by eliminating }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y^2=a(b-x^2)\quad\text{(1)}
\displaystyle \text{where }a\text{ and }b\text{ are parameters.}
\displaystyle \text{This equation contains two arbitrary constants, so we shall get a differential } \\ \text{equation of second order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2y\frac{dy}{dx}=-2ax\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle \left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}=-a\quad\text{(3)}
\displaystyle \text{From equations (2) and (3), we get}
\displaystyle y\frac{dy}{dx}=x\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 7: }~\text{Form the differential equation corresponding to } \\ y^{2}-2ay+x^{2}=a^{2}\text{ by eliminating }a. \hspace{5.0cm} \text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y^2-2ay+x^2=a^2\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{This equation contains only one arbitrary constant, so we shall get a differential } \\ \text{equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2y\frac{dy}{dx}-2a\frac{dy}{dx}+2x=0
\displaystyle \Rightarrow 2y\frac{dy}{dx}+2x=2a\frac{dy}{dx}
\displaystyle \Rightarrow a=y+\frac{x}{\frac{dy}{dx}}
\displaystyle \text{Substituting the value of }a\text{ in equation (1), we get}
\displaystyle y^2-2\left(y+\frac{x}{\frac{dy}{dx}}\right)y+x^2=\left(y+\frac{x}{\frac{dy}{dx}}\right)^2
\displaystyle \Rightarrow \frac{y^2\frac{dy}{dx}-2y\left(y\frac{dy}{dx}+x\right)+x^2\frac{dy}{dx}}{\frac{dy}{dx}}=\frac{\left(y\frac{dy}{dx}+x\right)^2}{\left(\frac{dy}{dx}\right)^2}
\displaystyle \Rightarrow y^2\left(\frac{dy}{dx}\right)^2-2y^2\left(\frac{dy}{dx}\right)^2-2xy\frac{dy}{dx}+x^2\left(\frac{dy}{dx}\right)^2=y^2\left(\frac{dy}{dx}\right)^2+2xy\frac{dy}{dx}+x^2
\displaystyle \Rightarrow (x^2-2y^2)\left(\frac{dy}{dx}\right)^2-4xy\frac{dy}{dx}-x^2=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 8: }~\text{Form the differential equation corresponding to } \\ (x-a)^{2}+(y-b)^{2}=r^{2}\text{ by eliminating }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }(x-a)^2+(y-b)^2=r^2\quad\text{(1)}
\displaystyle \text{where }a\text{ and }b\text{ are parameters.}
\displaystyle \text{This equation contains two parameters, so we shall get a second order } \\ \text{differential equation.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2(x-a)+2(y-b)\frac{dy}{dx}=0\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle 2+2\left(\frac{dy}{dx}\right)^2+2(y-b)\frac{d^2y}{dx^2}=0
\displaystyle \Rightarrow 1+\left(\frac{dy}{dx}\right)^2+(y-b)\frac{d^2y}{dx^2}=0
\displaystyle \Rightarrow (y-b)=-\frac{1+\left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}\quad\text{(3)}
\displaystyle \text{From equations (2) and (3), we get}
\displaystyle (x-a)-\frac{1+\left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}\frac{dy}{dx}=0
\displaystyle \Rightarrow (x-a)=\frac{\frac{dy}{dx}+\left(\frac{dy}{dx}\right)^3}{\frac{d^2y}{dx^2}}\quad\text{(4)}
\displaystyle \text{From equations (1), (3) and (4), we get}
\displaystyle \frac{\left[\frac{dy}{dx}+\left(\frac{dy}{dx}\right)^3\right]^2+\left[1+\left(\frac{dy}{dx}\right)^2\right]^2}{\left(\frac{d^2y}{dx^2}\right)^2}=r^2
\displaystyle \Rightarrow \frac{\left(\frac{dy}{dx}\right)^2+2\left(\frac{dy}{dx}\right)^4+\left(\frac{dy}{dx}\right)^6+1+2\left(\frac{dy}{dx}\right)^2+\left(\frac{dy}{dx}\right)^4}{\left(\frac{d^2y}{dx^2}\right)^2}=r^2
\displaystyle \Rightarrow 1+3\left(\frac{dy}{dx}\right)^2+3\left(\frac{dy}{dx}\right)^4+\left(\frac{dy}{dx}\right)^6=r^2\left(\frac{d^2y}{dx^2}\right)^2
\displaystyle \Rightarrow \left[1+\left(\frac{dy}{dx}\right)^2\right]^3=r^2\left(\frac{d^2y}{dx^2}\right)^2
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 9: }~\text{Find the differential equation of all the circles which pass } \\ \text{through the origin and whose centres lie on }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of circles that pass through the origin }(0,0)\text{ and whose centres } \\ \text{lie on the }y\text{-axis is }x^2+(y-a)^2=a^2\quad\text{(1)}
\displaystyle \text{where }a\text{ is an arbitrary constant.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a first order differential equation.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2x+2(y-a)\frac{dy}{dx}=0
\displaystyle \Rightarrow x+(y-a)\frac{dy}{dx}=0
\displaystyle \Rightarrow x=(a-y)\frac{dy}{dx}
\displaystyle \Rightarrow \frac{x}{\frac{dy}{dx}}=a-y
\displaystyle \Rightarrow a=y+\frac{x}{\frac{dy}{dx}}\quad\text{(2)}
\displaystyle \text{Substituting the value of }a\text{ from equation (2) in equation (1), we get}
\displaystyle x^2+\left(y-y-\frac{x}{\frac{dy}{dx}}\right)^2=\left(y+\frac{x}{\frac{dy}{dx}}\right)^2
\displaystyle \Rightarrow x^2+\frac{x^2}{\left(\frac{dy}{dx}\right)^2}=y^2+\frac{2xy}{\frac{dy}{dx}}+\frac{x^2}{\left(\frac{dy}{dx}\right)^2}
\displaystyle \Rightarrow x^2=y^2+\frac{2xy}{\frac{dy}{dx}}
\displaystyle \Rightarrow (x^2-y^2)\frac{dy}{dx}=2xy
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 10: }~\text{Find the differential equation of all the circles which pass } \\ \text{through the origin and whose centres lie on }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of circles that pass through the origin }(0,0)\text{ and whose } \\ \text{centres lie on the }x\text{-axis is }(x-a)^2+y^2=a^2\quad\text{(1)}
\displaystyle \text{where }a\text{ is an arbitrary constant.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a first order differential equation.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2(x-a)+2y\frac{dy}{dx}=0
\displaystyle \Rightarrow x-a+y\frac{dy}{dx}=0
\displaystyle \Rightarrow a=x+y\frac{dy}{dx}\quad\text{(2)}
\displaystyle \text{Substituting the value of }a\text{ from equation (2) in equation (1), we get}
\displaystyle (x-x-y\frac{dy}{dx})^2+y^2=\left(x+y\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow y^2\left(\frac{dy}{dx}\right)^2+y^2=x^2+2xy\frac{dy}{dx}+y^2\left(\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow 2xy\frac{dy}{dx}+x^2=y^2
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 11: }~\text{Assume that a rain drop evaporates at a rate proportional to its } \\ \text{surface area. Form a differential equation involving the rate of change of the radius } \\ \text{of the rain drop.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the surface area of the raindrop be }A.
\displaystyle \text{Thus, the rate of evaporation will be given by }\frac{dV}{dt}.
\displaystyle \text{As per the given condition,}
\displaystyle \frac{dV}{dt}\propto A.
\displaystyle \Rightarrow \frac{dV}{dt}=-kA.
\displaystyle \text{Here, }k\text{ is a constant. Also, the negative sign appears since }V\text{ decreases as }t\text{ increases.}
\displaystyle \text{Now, }V=\frac{4}{3}\pi r^3.
\displaystyle \text{Here, }r\text{ is the radius of the spherical drop.}
\displaystyle \therefore \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=-k\times4\pi r^2.
\displaystyle \Rightarrow \frac{4}{3}\times3\pi r^2\frac{dr}{dt}=-k\times4\pi r^2.
\displaystyle \Rightarrow \frac{dr}{dt}=-k.
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 12: }~\text{Find the differential equation of all the parabolas with latus } \\ \text{rectum }4a\text{ and whose axes are parallel to }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of parabolas with latus rectum }4a\text{ and axis } \\ \text{parallel to the }x\text{-axis is }(y-\beta)^2=4a(x-\alpha)\quad\text{(1)}
\displaystyle \text{where }\alpha\text{ and }\beta\text{ are two arbitrary constants.}
\displaystyle \text{As this equation has two arbitrary constants, we shall get a second order differential equation.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2(y-\beta)\frac{dy}{dx}=4a\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle 2\left(\frac{dy}{dx}\right)^2+2(y-\beta)\frac{d^2y}{dx^2}=0\quad\text{(3)}
\displaystyle \text{Now, from equation (2), we get }y-\beta=\frac{2a}{\frac{dy}{dx}}\quad\text{(4)}
\displaystyle \text{Substituting from (4) into (3), we get}
\displaystyle 2\left(\frac{dy}{dx}\right)^2+2\left(\frac{2a}{\frac{dy}{dx}}\right)\frac{d^2y}{dx^2}=0
\displaystyle \Rightarrow 2a\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^3=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 13: }~\text{Show that the differential equation of which }y=2(x^{2}-1)+ce^{-x^{2}} \\ \text{ is a solution, is }\frac{dy}{dx}+2xy=4x^{3}.
\displaystyle \text{Answer:}
\displaystyle \text{The given equation is }y=2(x^2-1)+ce^{-x^2}\quad\text{(1)}
\displaystyle \text{where }c\text{ is a parameter.}
\displaystyle \text{As this equation has one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{dy}{dx}=4x+c(-2x)e^{-x^2}
\displaystyle \Rightarrow \frac{dy}{dx}=4x-2x\,ce^{-x^2}\quad\text{(2)}
\displaystyle \text{From equation (1), we get }ce^{-x^2}=y-2(x^2-1)
\displaystyle \Rightarrow ce^{-x^2}=y-2x^2+2
\displaystyle \text{Substituting this in equation (2), we get}
\displaystyle \frac{dy}{dx}=4x-2x(y-2x^2+2)
\displaystyle \Rightarrow \frac{dy}{dx}=4x-2xy+4x^3-4x
\displaystyle \Rightarrow \frac{dy}{dx}+2xy=4x^3
\displaystyle \text{Hence, }y=2(x^2-1)+ce^{-x^2}\text{ is the solution of the differential equation }\frac{dy}{dx}+2xy=4x^3.

\displaystyle \textbf{Question 14: }~\text{Form the differential equation having } y=(\sin^{-1}x)^{2}+A\cos^{-1}x+B, \text{ where }A\text{ and }B\text{ are arbitrary constants, as its general solution.}
\displaystyle \text{Answer:}
\displaystyle \text{We have two constants in the solution, so we shall differentiate both sides } \\ \text{twice and eliminate the constants }A\text{ and }B.
\displaystyle y=(\sin^{-1}x)^2+A\cos^{-1}x+B
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2\sin^{-1}x}{\sqrt{1-x^2}}-\frac{A}{\sqrt{1-x^2}}
\displaystyle \Rightarrow \sqrt{1-x^2}\,\frac{dy}{dx}=2\sin^{-1}x-A
\displaystyle \Rightarrow \frac{d}{dx}\!\left(\sqrt{1-x^2}\,\frac{dy}{dx}\right)=\frac{2}{\sqrt{1-x^2}}
\displaystyle \Rightarrow \sqrt{1-x^2}\,\frac{d^2y}{dx^2}-\frac{x}{\sqrt{1-x^2}}\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}
\displaystyle \Rightarrow (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2
\displaystyle \Rightarrow (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2=0

\displaystyle \textbf{Question 15: }~\text{Form the differential equation of the family of curves } \\ \text{represented by the equation (}a\text{ being the parameter):}
\displaystyle \text{(i): }(2x+a)^{2}+y^{2}=a^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }(2x+a)^2+y^2=a^2\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one parameter, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2(2x+a)\cdot2+2y\frac{dy}{dx}=0\quad\text{(2)}
\displaystyle \text{Now, from equation (1), we get}
\displaystyle 4x^2+4ax+a^2+y^2=a^2
\displaystyle \Rightarrow 4ax=-y^2-4x^2
\displaystyle \Rightarrow a=-\frac{4x^2+y^2}{4x}
\displaystyle \text{Putting the value of }a\text{ in equation (2), we get}
\displaystyle 4\left(2x-\frac{4x^2+y^2}{4x}\right)+2y\frac{dy}{dx}=0
\displaystyle \Rightarrow 4\left(\frac{8x^2-4x^2-y^2}{4x}\right)+2y\frac{dy}{dx}=0
\displaystyle \Rightarrow 4x^2-y^2+2xy\frac{dy}{dx}=0
\displaystyle \Rightarrow y^2-4x^2-2xy\frac{dy}{dx}=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(ii): }(2x-a)^{2}-y^{2}=a^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }(2x-a)^2-y^2=a^2
\displaystyle \Rightarrow 4x^2-4ax+a^2-y^2=a^2
\displaystyle \Rightarrow 4x^2-4ax-y^2=0\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one parameter, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 8x-4a-2y\frac{dy}{dx}=0
\displaystyle \Rightarrow -y\frac{dy}{dx}+4x=2a\quad\text{(2)}
\displaystyle \text{Now, from equation (1), we get}
\displaystyle 2a=\frac{4x^2-y^2}{2x}\quad\text{(3)}
\displaystyle \text{From equations (2) and (3), we get}
\displaystyle -y\frac{dy}{dx}+4x=\frac{4x^2-y^2}{2x}
\displaystyle \Rightarrow -2xy\frac{dy}{dx}+8x^2=4x^2-y^2
\displaystyle \Rightarrow -2xy\frac{dy}{dx}+4x^2+y^2=0
\displaystyle \Rightarrow 2xy\frac{dy}{dx}=4x^2+y^2
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(iii): }(x-a)^{2}+2y^{2}=a^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }(x-a)^2+2y^2=a^2
\displaystyle \Rightarrow x^2-2ax+a^2+2y^2=a^2
\displaystyle \Rightarrow x^2-2ax+2y^2=0\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2x-2a+4y\frac{dy}{dx}=0\quad\text{(2)}
\displaystyle \text{From equation (1), we get}
\displaystyle 2a=\frac{x^2+2y^2}{x}\quad\text{(3)}
\displaystyle \text{From equations (2) and (3), we get}
\displaystyle 2x-\frac{x^2+2y^2}{x}+4y\frac{dy}{dx}=0
\displaystyle \Rightarrow 2x^2-x^2-2y^2+4xy\frac{dy}{dx}=0
\displaystyle \Rightarrow x^2-2y^2+4xy\frac{dy}{dx}=0
\displaystyle \Rightarrow 4xy\frac{dy}{dx}=2y^2-x^2
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 16: }~\text{Represent the following families of curves by forming the } \\ \text{corresponding differential equations (}a,b\text{ being parameters):}
\displaystyle \text{(i): }x^{2}+y^{2}=a^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }x^2+y^2=a^2\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2x+2y\frac{dy}{dx}=0
\displaystyle \Rightarrow x+y\frac{dy}{dx}=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(ii): }x^{2}-y^{2}=a^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }x^2-y^2=a^2\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2x-2y\frac{dy}{dx}=0
\displaystyle \Rightarrow x-y\frac{dy}{dx}=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(iii): }y^{2}=4ax.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y^2=4ax\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2y\frac{dy}{dx}=4a
\displaystyle \Rightarrow 2y\frac{dy}{dx}=\frac{y^2}{x}\quad\text{[Using equation (1)]}
\displaystyle \Rightarrow 2x\frac{dy}{dx}=y
\displaystyle \Rightarrow y-2x\frac{dy}{dx}=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(iv): }x^{2}+(y-b)^{2}=1.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }x^2+(y-b)^2=1\quad\text{(1)}
\displaystyle \text{where }b\text{ is a parameter.}
\displaystyle \text{As this equation contains only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2x+2(y-b)\frac{dy}{dx}=0
\displaystyle \Rightarrow 2x+2\sqrt{1-x^2}\frac{dy}{dx}=0\quad\text{[Using equation (1)]}
\displaystyle \Rightarrow x=-\sqrt{1-x^2}\frac{dy}{dx}
\displaystyle \Rightarrow x^2=(1-x^2)\left(\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow x^2=\left(\frac{dy}{dx}\right)^2-x^2\left(\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow x^2\left[1+\left(\frac{dy}{dx}\right)^2\right]=\left(\frac{dy}{dx}\right)^2
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(v): }(x-a)^{2}-y^{2}=1.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }(x-a)^2-y^2=1\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2(x-a)-2y\frac{dy}{dx}=0
\displaystyle \Rightarrow (x-a)-y\frac{dy}{dx}=0
\displaystyle \Rightarrow x-a=y\frac{dy}{dx}
\displaystyle \Rightarrow \sqrt{1+y^2}=y\frac{dy}{dx}\quad\text{[Using equation (1)]}
\displaystyle \Rightarrow 1+y^2=y^2\left(\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow y^2\left(\frac{dy}{dx}\right)^2-y^2=1
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(vi): }\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\quad\text{(1)}
\displaystyle \text{where }a\text{ and }b\text{ are parameters.}
\displaystyle \text{As this equation has two arbitrary constants, we shall get a differential equation of second order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle \frac{2}{a^2}-\frac{2}{b^2}\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]=0
\displaystyle \Rightarrow \frac{2}{a^2}=\frac{2}{b^2}\left[y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right]
\displaystyle \Rightarrow \frac{b^2}{a^2}=y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\quad\text{(3)}
\displaystyle \text{Now, from equation (2), we get}
\displaystyle \frac{2x}{a^2}=\frac{2y}{b^2}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{b^2}{a^2}=\frac{y}{x}\frac{dy}{dx}\quad\text{(4)}
\displaystyle \text{From equations (3) and (4), we get}
\displaystyle \frac{y}{x}\frac{dy}{dx}=y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow x\left[y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right]=y\frac{dy}{dx}
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(vii): }y^{2}=4a(x-b).
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y^2=4a(x-b)\quad\text{(1)}
\displaystyle \text{where }a\text{ and }b\text{ are parameters.}
\displaystyle \text{As this equation has two arbitrary constants, we shall get a differential equation of second order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2y\frac{dy}{dx}=4a
\displaystyle \Rightarrow y\frac{dy}{dx}=2a\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(viii): }y=ax^{3}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y=ax^3\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{dy}{dx}=3ax^2
\displaystyle \Rightarrow \frac{dy}{dx}=3\frac{y}{x^3}\,x^2\quad\text{[Using equation (1)]}
\displaystyle \Rightarrow x\frac{dy}{dx}=3y
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(ix): }x^{2}+y^{2}=ax^{3}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }x^2+y^2=ax^3\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2x+2y\frac{dy}{dx}=3ax^2
\displaystyle \Rightarrow 2x+2y\frac{dy}{dx}=3\left(\frac{x^2+y^2}{x^3}\right)x^2\quad\text{[Using equation (1)]}
\displaystyle \Rightarrow 2x+2y\frac{dy}{dx}=3\frac{x^2+y^2}{x}
\displaystyle \Rightarrow 2x^2+2xy\frac{dy}{dx}=3x^2+3y^2
\displaystyle \Rightarrow 2xy\frac{dy}{dx}=x^2+3y^2
\displaystyle \text{It is the required differential equation.}

\displaystyle \text{(x): }y=e^{ax}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is }y=e^{ax}
\displaystyle \Rightarrow \log y=ax\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation has only one arbitrary constant, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{1}{y}\frac{dy}{dx}=a
\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{\log y}{x}\quad\text{[Using equation (1)]}
\displaystyle \Rightarrow x\frac{dy}{dx}=y\log y
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 17: }~\text{Form the differential equation representing the family of } \\ \text{ellipses having centre at the origin and foci on }x\text{-axis.} \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of ellipses having centre at the origin and foci on the } \\ x\text{-axis is }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\quad\text{(1)}
\displaystyle \text{where }a\text{ and }b\text{ are parameters.}
\displaystyle \text{As this equation contains two parameters, we shall get a second-order differential equation.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle \frac{2}{a^2}+\frac{2}{b^2}\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]=0
\displaystyle \Rightarrow \frac{2}{a^2}=-\frac{2}{b^2}\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]
\displaystyle \Rightarrow \frac{b^2}{a^2}=-\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]\quad\text{(3)}
\displaystyle \text{Now, from equation (2), we get}
\displaystyle \frac{x}{a^2}=-\frac{y}{b^2}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{b^2}{a^2}=-\frac{y}{x}\frac{dy}{dx}\quad\text{(4)}
\displaystyle \text{From equations (3) and (4), we get}
\displaystyle -\frac{y}{x}\frac{dy}{dx}=-\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]
\displaystyle \Rightarrow \frac{y}{x}\frac{dy}{dx}=\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}
\displaystyle \Rightarrow y\frac{dy}{dx}=x\left(\frac{dy}{dx}\right)^2+xy\frac{d^2y}{dx^2}
\displaystyle \Rightarrow xy\frac{d^2y}{dx^2}+x\left(\frac{dy}{dx}\right)^2-y\frac{dy}{dx}=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 18: }~\text{Form the differential equation of the family of hyperbolas } \\ \text{having foci on }x\text{-axis and centre at the origin.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of hyperbolas having the centre at the origin and foci on the } \\ x\text{-axis is }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\quad\text{(1)}
\displaystyle \text{where }a\text{ and }b\text{ are parameters.}
\displaystyle \text{As this equation contains two parameters, we shall get a second-order differential equation.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0\quad\text{(2)}
\displaystyle \text{Differentiating equation (2) with respect to }x\text{, we get}
\displaystyle \frac{2}{a^2}-\frac{2}{b^2}\left[y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right]=0
\displaystyle \Rightarrow \frac{1}{a^2}=\frac{1}{b^2}\left[y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right]
\displaystyle \Rightarrow \frac{b^2}{a^2}=y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\quad\text{(3)}
\displaystyle \text{Now, from equation (2), we get}
\displaystyle \frac{2x}{a^2}=\frac{2y}{b^2}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{b^2}{a^2}=\frac{y}{x}\frac{dy}{dx}\quad\text{(4)}
\displaystyle \text{From equations (3) and (4), we get}
\displaystyle \frac{y}{x}\frac{dy}{dx}=y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow y\frac{dy}{dx}=xy\frac{d^2y}{dx^2}+x\left(\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow xy\frac{d^2y}{dx^2}+x\left(\frac{dy}{dx}\right)^2-y\frac{dy}{dx}=0
\displaystyle \text{It is the required differential equation.}

\displaystyle \textbf{Question 19: }~\text{Form the differential equation of the family of circles in } \\ \text{the second quadrant and touching the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of circles in the second quadrant and touching the } \\ \text{coordinate axes is }(x+a)^2+(y-a)^2=a^2
\displaystyle \Rightarrow x^2+2ax+a^2+y^2-2ay+a^2=a^2
\displaystyle \Rightarrow x^2+2ax+y^2-2ay+a^2=0\quad\text{(1)}
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{As this equation contains one parameter, we shall get a differential equation of first order.}
\displaystyle \text{Differentiating equation (1) with respect to }x\text{, we get}
\displaystyle 2x+2a+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0
\displaystyle \Rightarrow x+y\frac{dy}{dx}+a\left(1-\frac{dy}{dx}\right)=0
\displaystyle \Rightarrow a=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}\quad\text{(2)}
\displaystyle \text{Substituting the value of }a\text{ from (2) in (1), we get}
\displaystyle x^2+2x\!\left(\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}\right)+y^2-2y\!\left(\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}\right)+\left(\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}\right)^2=0
\displaystyle \Rightarrow (x+y\frac{dy}{dx})^2=(x+y)^2\!\left[1+\left(\frac{dy}{dx}\right)^2\right]
\displaystyle \Rightarrow (x+y\frac{dy}{dx})^2-(x+y)^2\!\left[1+\left(\frac{dy}{dx}\right)^2\right]=0
\displaystyle \text{It is the required differential equation.}

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