\displaystyle \textbf{Question 1: }\ \ \text{If } y=A\cos nx+B\sin nx,\ \text{show that } \\ \frac{d^2y}{dx^2}+n^2y=0. \hspace{5.0cm} \text{[CBSE 2001C]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle y=A\cos nx+B\sin nx
\displaystyle \text{On differentiating with respect to } x,\ \text{we get}
\displaystyle \frac{dy}{dx}=-An\sin nx+Bn\cos nx
\displaystyle \text{On differentiating again with respect to } x,\ \text{we get}
\displaystyle \frac{d^2y}{dx^2}=-An^2\cos nx-Bn^2\sin nx=-n^2(A\cos nx+B\sin nx)=-n^2y
\displaystyle \therefore\ \frac{d^2y}{dx^2}+n^2y=0.

\displaystyle \textbf{Question 2: }\ \ \text{If } y=Ae^{mx}+Be^{nx},\ \text{show that } \\ \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=0. \hspace{5.0cm} \text{[CBSE 2007, 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have, } y=Ae^{mx}+Be^{nx}
\displaystyle \therefore\ \frac{dy}{dx}=Ame^{mx}+Bne^{nx}
\displaystyle \Rightarrow\ \frac{d^2y}{dx^2}=Am^2e^{mx}+Bn^2e^{nx}
\displaystyle \therefore\ \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny
\displaystyle =(Am^2e^{mx}+Bn^2e^{nx})-(m+n)(Ame^{mx}+Bne^{nx})+mn(Ae^{mx}+Be^{nx})=0.

\displaystyle \textbf{Question 3: }\ \ \text{If } y=A\cos(\log x)+B\sin(\log x),\ \text{prove that } \\ x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0. \hspace{5.0cm} \text{[CBSE 2007,2009]}
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle y=A\cos(\log x)+B\sin(\log x)
\displaystyle \text{On differentiating with respect to } x,\ \text{we get}
\displaystyle \frac{dy}{dx}=-\frac{1}{x}A\sin(\log x)+\frac{B}{x}\cos(\log x)
\displaystyle \Rightarrow\ x\frac{dy}{dx}=-A\sin(\log x)+B\cos(\log x)
\displaystyle \text{On differentiating again with respect to } x,\ \text{we get}
\displaystyle x\frac{d^2y}{dx^2}+\frac{dy}{dx}=-\frac{A\cos(\log x)}{x}-\frac{B\sin(\log x)}{x}
\displaystyle \Rightarrow\ x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}=-(A\cos(\log x)+B\sin(\log x))
\displaystyle \Rightarrow\ x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}=-y
\displaystyle \Rightarrow\ x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0

\displaystyle \textbf{Question 4: }\ \ \text{If } (ax+b)e^{\frac{y}{x}}=x\ \text{or, } y=x\log\left(\frac{x}{a+bx}\right),\ \text{prove that } \\  x^3\frac{d^2y}{dx^2}=\left(x\frac{dy}{dx}-y\right)^2. \hspace{5.0cm} \text{[CBSE 2005, 2013, 2015]}
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle (ax+b)e^{\frac{y}{x}}=x
\displaystyle \Rightarrow\ e^{\frac{y}{x}}=\frac{x}{ax+b}
\displaystyle \Rightarrow\ \frac{y}{x}=\log\left(\frac{x}{ax+b}\right)
\displaystyle \Rightarrow\ y=x\log\left(\frac{x}{a+bx}\right)
\displaystyle \Rightarrow\ y=x\{\log x-\log(a+bx)\}
\displaystyle \Rightarrow\ \frac{y}{x}=\log x-\log(a+bx)
\displaystyle \text{On differentiating with respect to } x,\ \text{we get}
\displaystyle \frac{x\frac{dy}{dx}-y}{x^2}=\frac{1}{x}-\frac{1}{a+bx}\frac{d}{dx}(a+bx)=\frac{1}{x}-\frac{b}{a+bx}
\displaystyle \Rightarrow\ x\frac{dy}{dx}-y=x^2\left\{\frac{1}{x}-\frac{b}{a+bx}\right\}
\displaystyle \Rightarrow\ x\frac{dy}{dx}-y=\frac{ax}{a+bx}\ \ \ (i)
\displaystyle \text{Differentiating both sides of (i) with respect to } x,\ \text{we get}
\displaystyle x\frac{d^2y}{dx^2}+\frac{dy}{dx}-\frac{dy}{dx}=\frac{(a+bx)a-ax(0+b)}{(a+bx)^2}
\displaystyle \Rightarrow\ x\frac{d^2y}{dx^2}=\frac{a^2}{(a+bx)^2}
\displaystyle \Rightarrow\ x^3\frac{d^2y}{dx^2}=\frac{a^2x^2}{(a+bx)^2}\ \ \ \text{[Multiplying both sides by } x^2]
\displaystyle \Rightarrow\ x^3\frac{d^2y}{dx^2}=\left(\frac{ax}{a+bx}\right)^2\ \ \ (ii)
\displaystyle \text{From (i) and (ii), we obtain}
\displaystyle x^3\frac{d^2y}{dx^2}=\left(x\frac{dy}{dx}-y\right)^2.

\displaystyle \textbf{Question 5: }\ \ \text{If } y=\log\left\{x+\sqrt{x^2+a^2}\right\},\ \text{prove that: } \\ (x^2+a^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0. \hspace{5.0cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle y=\log\left\{x+\sqrt{x^2+a^2}\right\}
\displaystyle \text{On differentiating with respect to } x,\ \text{we get}
\displaystyle \frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\times\frac{d}{dx}\left\{x+\sqrt{x^2+a^2}\right\}=\frac{1}{x+\sqrt{x^2+a^2}}\left\{1+\frac{2x}{2\sqrt{x^2+a^2}}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\times\left\{\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\right\}
\displaystyle \Rightarrow\ y_1=\frac{1}{\sqrt{x^2+a^2}},\ \text{where } y_1=\frac{dy}{dx}
\displaystyle \Rightarrow\ y_1^2(x^2+a^2)=1
\displaystyle \text{Differentiating with respect to } x,\ \text{we get}
\displaystyle y_1^2\frac{d}{dx}(x^2+a^2)+(x^2+a^2)\frac{d}{dx}(y_1^2)=0
\displaystyle \Rightarrow\ y_1^2(2x)+(x^2+a^2)\times2y_1y_2=0\ \ \left[\because\ \frac{d}{dx}(y_1^2)=2(y_1)^{2-1}\frac{d}{dx}(y_1)=2y_1\frac{d}{dx}\left(\frac{dy}{dx}\right)=2y_1y_2\right]
\displaystyle \Rightarrow\ 2y_1\left\{y_2(x^2+a^2)+xy_1\right\}=0
\displaystyle \Rightarrow\ y_2(x^2+a^2)+xy_1=0\ \ \left[\because\ y_1\neq0\right]
\displaystyle \Rightarrow\ (x^2+a^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0

\displaystyle \textbf{Question 6: }\ \ \text{If } y=\sin^{-1}x,\ \text{then show that }(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=0.\ \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle y=\sin^{-1}x
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\ \ \ \text{[Differentiating with respect to }x]
\displaystyle \Rightarrow\ \sqrt{1-x^2}\ \frac{dy}{dx}=1
\displaystyle \text{Differentiating both sides with respect to }x,\ \text{we get}
\displaystyle \sqrt{1-x^2}\ \frac{d^2y}{dx^2}-\frac{x}{\sqrt{1-x^2}}\ \frac{dy}{dx}=0
\displaystyle \Rightarrow\ (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=0\ \ \ \text{[Multiplying both sides by }\sqrt{1-x^2}]
\displaystyle \textbf{ALTER}\ \ \text{We have,}
\displaystyle y=\sin^{-1}x
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}
\displaystyle \Rightarrow\ (1-x^2)\left(\frac{dy}{dx}\right)^2=1
\displaystyle \text{Differentiating both sides with respect to }x,\ \text{we get}
\displaystyle (1-x^2)\left\{2\frac{dy}{dx}\times\frac{d}{dx}\left(\frac{dy}{dx}\right)\right\}-2x\left(\frac{dy}{dx}\right)^2=0
\displaystyle \Rightarrow\ 2(1-x^2)\frac{dy}{dx}\frac{d^2y}{dx^2}-2x\left(\frac{dy}{dx}\right)^2=0
\displaystyle \Rightarrow\ (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=0

\displaystyle \textbf{Question 7: }\ \ \text{If } y=e^{m\sin^{-1}x},\ \text{prove that } \\ (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-m^2y=0. \hspace{5.0cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle y=e^{m\sin^{-1}x}
\displaystyle \text{Differentiating with respect to }x,\ \text{we obtain}
\displaystyle \frac{dy}{dx}=e^{m\sin^{-1}x}\times\frac{m}{\sqrt{1-x^2}}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{my}{\sqrt{1-x^2}}\ \ \ \left[\because\ e^{m\sin^{-1}x}=y\right]
\displaystyle \Rightarrow\ \left(\frac{dy}{dx}\right)^2=\frac{m^2y^2}{1-x^2}
\displaystyle \Rightarrow\ (1-x^2)\left(\frac{dy}{dx}\right)^2=m^2y^2
\displaystyle \Rightarrow\ (1-x^2)y_1^2=m^2y^2,\ \text{where } y_1=\frac{dy}{dx}
\displaystyle \text{Differentiating with respect to }x,\ \text{we obtain}
\displaystyle (1-x^2)\frac{d}{dx}(y_1^2)+y_1^2\frac{d}{dx}(1-x^2)=m^2\frac{d}{dx}(y^2)
\displaystyle \Rightarrow\ (1-x^2)2y_1y_2+y_1^2(-2x)=m^2(2yy_1)\ \ \ \left[\because\ \frac{d}{dx}(y_1^2)=2y_1y_2\ \text{and}\ \frac{d}{dx}(y^2)=2yy_1\right]
\displaystyle \Rightarrow\ 2y_1\left\{(1-x^2)y_2-xy_1-m^2y\right\}=0
\displaystyle \Rightarrow\ (1-x^2)y_2-xy_1-m^2y=0\ \ \ \left[\because\ y_1\neq0\right]
\displaystyle \Rightarrow\ (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-m^2y=0

\displaystyle \textbf{Question 8: }\ \ \text{If } y=\left\{x+\sqrt{x^2+1}\right\}^m,\ \text{show that } \\ (x^2+1)y_2+xy_1-m^2y=0. \hspace{5.0cm} \text{[CBSE 2013, 2015]}
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle y=\left\{x+\sqrt{x^2+1}\right\}^m
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle \frac{dy}{dx}=m\left\{x+\sqrt{x^2+1}\right\}^{m-1}\times\frac{d}{dx}\left\{x+\sqrt{x^2+1}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=m\left\{x+\sqrt{x^2+1}\right\}^{m-1}\times\left\{1+\frac{2x}{2\sqrt{x^2+1}}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{m\left\{\sqrt{x^2+1}+x\right\}^m}{\sqrt{x^2+1}}=\frac{my}{\sqrt{x^2+1}}
\displaystyle \Rightarrow\ y_1=\frac{my}{\sqrt{x^2+1}}
\displaystyle \Rightarrow\ y_1\sqrt{x^2+1}=my
\displaystyle \Rightarrow\ y_1^2(x^2+1)=m^2y^2\ \ \ \text{[Squaring both sides]}
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle 2y_1y_2(1+x^2)+y_1^2(2x)=2m^2yy_1
\displaystyle \Rightarrow\ y_2(1+x^2)+xy_1-m^2y=0

\displaystyle \textbf{Question 9: }\ \ \text{If } y=\frac{\sin^{-1}x}{\sqrt{1-x^2}},\ \text{show that } \\ (1-x^2)\frac{d^2y}{dx^2}-3x\frac{dy}{dx}-y=0. \hspace{5.0cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle y=\frac{\sin^{-1}x}{\sqrt{1-x^2}}
\displaystyle \Rightarrow\ y\sqrt{1-x^2}=\sin^{-1}x
\displaystyle \text{Differentiating both sides with respect to }x,\ \text{we get}
\displaystyle \frac{dy}{dx}\sqrt{1-x^2}-\frac{x}{\sqrt{1-x^2}}y=\frac{1}{\sqrt{1-x^2}}
\displaystyle \Rightarrow\ \frac{dy}{dx}(1-x^2)-xy=1
\displaystyle \text{Differentiating both sides with respect to }x,\ \text{we get}
\displaystyle \frac{d^2y}{dx^2}(1-x^2)-2x\frac{dy}{dx}-x\frac{dy}{dx}-y=0
\displaystyle \Rightarrow\ (1-x^2)\frac{d^2y}{dx^2}-3x\frac{dy}{dx}-y=0

\displaystyle \textbf{Question 10: }\ \ \text{If } x=\tan\left(\frac{1}{a}\log y\right),\ \text{show that } \\ (1+x^2)\frac{d^2y}{dx^2}+(2x-a)\frac{dy}{dx}=0. \hspace{5.0cm} \text{[CBSE 2011, 2013]}
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle x=\tan\left(\frac{1}{a}\log y\right)
\displaystyle \Rightarrow\ \tan^{-1}x=\frac{1}{a}\log y
\displaystyle \Rightarrow\ a\tan^{-1}x=\log y
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle \frac{a}{1+x^2}=\frac{1}{y}\frac{dy}{dx}
\displaystyle \Rightarrow\ (1+x^2)\frac{dy}{dx}=ay
\displaystyle \text{Differentiating with respect to }x
\displaystyle (1+x^2)\frac{d^2y}{dx^2}+2x\frac{dy}{dx}=a\frac{dy}{dx}
\displaystyle \Rightarrow\ (1+x^2)\frac{d^2y}{dx^2}+(2x-a)\frac{dy}{dx}=0

\displaystyle \textbf{Question 11: }\ \ \text{If } y=x^x,\ \text{prove that } \\ \frac{d^2y}{dx^2}-\frac{1}{y}\left(\frac{dy}{dx}\right)^2-\frac{y}{x}=0. \hspace{6.0cm} \text{[CBSE 2014, 2016]}
\displaystyle \text{Answer:}
\displaystyle   \text{We have, } y=x^x
\displaystyle \text{or, } y=e^{\log x^x}=e^{x\log x}
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle \frac{dy}{dx}=e^{x\log x}\frac{d}{dx}(x\log x)
\displaystyle \Rightarrow\ \frac{dy}{dx}=x^x(1+\log x)
\displaystyle \Rightarrow\ \frac{dy}{dx}=y(1+\log x)\ \ \ (i)
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle \frac{d^2y}{dx^2}=y\frac{d}{dx}(1+\log x)+\frac{dy}{dx}(1+\log x)
\displaystyle \Rightarrow\ \frac{d^2y}{dx^2}=\frac{y}{x}+\frac{dy}{dx}(1+\log x)
\displaystyle \Rightarrow\ \frac{d^2y}{dx^2}=\frac{y}{x}+\frac{dy}{dx}\left(\frac{1}{y}\frac{dy}{dx}\right)\ \ \ \left[\text{From (i), }1+\log x=\frac{1}{y}\frac{dy}{dx}\right]
\displaystyle \Rightarrow\ \frac{d^2y}{dx^2}=\frac{y}{x}+\frac{1}{y}\left(\frac{dy}{dx}\right)^2
\displaystyle \Rightarrow\ \frac{d^2y}{dx^2}-\frac{1}{y}\left(\frac{dy}{dx}\right)^2-\frac{y}{x}=0

\displaystyle \textbf{Question 12: }\ \ \text{If } x=a\cos^3\theta,\ y=a\sin^3\theta,\ \text{find } \frac{d^2y}{dx^2}. \\ \text{Also, find its value at } \theta=\frac{\pi}{6}. \hspace{6.0cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle   \text{We have, } x=a\cos^3\theta \text{ and } y=a\sin^3\theta
\displaystyle \frac{dx}{d\theta}=-3a\cos^2\theta\sin\theta \text{ and } \frac{dy}{d\theta}=3a\sin^2\theta\cos\theta
\displaystyle \text{So, } \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}=-\tan\theta
\displaystyle \text{Differentiating both sides with respect to }x,\ \text{we obtain,}
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(-\tan\theta)=-\sec^2\theta\frac{d\theta}{dx}=-\sec^2\theta\times\frac{1}{-3a\cos^2\theta\sin\theta}=\frac{1}{3a}\sec^4\theta\mathrm{cosec}\theta
\displaystyle \therefore\ \left(\frac{d^2y}{dx^2}\right)_{\theta=\frac{\pi}{6}}=\frac{1}{3a}\sec^4\frac{\pi}{6}\mathrm{cosec}\frac{\pi}{6}=\frac{1}{3a}\times\left(\frac{2}{\sqrt{3}}\right)^4\times2=\frac{32}{27a}

\displaystyle \textbf{Question 13: }\ \ \text{If } x=a\sin t \text{ and } y=a\left(\cos t+\log\tan\frac{t}{2}\right),\ \text{find } \frac{d^2y}{dx^2}.\ \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, } x=a\sin t \text{ and } y=a\left(\cos t+\log\tan\frac{t}{2}\right)
\displaystyle \frac{dx}{dt}=a\cos t \text{ and } \frac{dy}{dt}=a\left(-\sin t+\frac{1}{\tan\frac{t}{2}}\times\sec^2\frac{t}{2}\times\frac{1}{2}\right)
\displaystyle \Rightarrow\ \frac{dx}{dt}=a\cos t \text{ and } \frac{dy}{dt}=a\left(-\sin t+\frac{1}{\sin t}\right)
\displaystyle \Rightarrow\ \frac{dx}{dt}=a\cos t \text{ and } \frac{dy}{dt}=\frac{a(1-\sin^2 t)}{\sin t}
\displaystyle \Rightarrow\ \frac{dx}{dt}=a\cos t \text{ and } \frac{dy}{dt}=\frac{a\cos^2 t}{\sin t}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos^2 t/\sin t}{a\cos t}=\frac{\cos t}{\sin t}=\cot t
\displaystyle \Rightarrow\ \frac{d^2y}{dx^2}=\frac{d}{dx}(\cot t)=-\mathrm{cosec}^2 t\frac{dt}{dx}
\displaystyle \Rightarrow\ \frac{d^2y}{dx^2}=-\mathrm{cosec}^2 t\times\frac{1}{a\cos t}=-\frac{1}{a\sin^2 t\cos t}

\displaystyle \textbf{Question 14: }\ \ \text{If } x=\sin t \text{ and } y=\sin pt,\ \text{prove that } \\ (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0. \hspace{8.0cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle    \text{We have,}
\displaystyle x=\sin t,\ y=\sin pt\ \Rightarrow\ \frac{dx}{dt}=\cos t \text{ and } \frac{dy}{dt}=p\cos pt
\displaystyle \therefore\ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{p\cos pt}{\cos t}=\frac{p\sqrt{1-\sin^2 pt}}{\sqrt{1-\sin^2 t}}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{p\sqrt{1-y^2}}{\sqrt{1-x^2}}
\displaystyle \Rightarrow\ \left(\frac{dy}{dx}\right)^2(1-x^2)=p^2(1-y^2)\ \ \ \text{[Squaring both sides]}
\displaystyle \text{Differentiating with respect to }x,\ \text{we obtain}
\displaystyle (1-x^2)\frac{d}{dx}\left(\frac{dy}{dx}\right)^2+\left(\frac{dy}{dx}\right)^2\frac{d}{dx}(1-x^2)=p^2\left(0-2y\frac{dy}{dx}\right)
\displaystyle \Rightarrow\ (1-x^2)2\frac{dy}{dx}\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2(-2x)=-2p^2y\frac{dy}{dx}
\displaystyle \Rightarrow\ 2\frac{dy}{dx}\left\{(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y\right\}=0
\displaystyle \Rightarrow\ (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0\ \ \ \left[\because\ 2\frac{dy}{dx}\neq0\right]

\displaystyle \textbf{Question 15: }\ \ \text{If } x=a\cos\theta+b\sin\theta \text{ and } y=a\sin\theta-b\cos\theta,\ \text{prove that } \\  y^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=0. \hspace{5.0cm} \text{[CBSE 2013, 2014, 2015]}

\displaystyle \text{Answer:}
\displaystyle  \text{We have, } x=a\cos\theta+b\sin\theta \text{ and } y=a\sin\theta-b\cos\theta
\displaystyle \Rightarrow\ x^2+y^2=(a\cos\theta+b\sin\theta)^2+(a\sin\theta-b\cos\theta)^2
\displaystyle \Rightarrow\ x^2+y^2=a^2(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta)
\displaystyle \Rightarrow\ x^2+y^2=a^2+b^2
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle 2x+2y\frac{dy}{dx}=0
\displaystyle \Rightarrow\ \frac{dy}{dx}=-\frac{x}{y}\ \ \ (i)
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle \frac{d^2y}{dx^2}=-\left\{\frac{y\times1-x\frac{dy}{dx}}{y^2}\right\}=-\left\{\frac{y-x\left(-\frac{x}{y}\right)}{y^2}\right\}\ \ \ \text{[Using (i)]}
\displaystyle \Rightarrow\ \frac{d^2y}{dx^2}=-\frac{x^2+y^2}{y^3}\ \ \ (ii)
\displaystyle \therefore\ y^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=-y^2\left(\frac{x^2+y^2}{y^3}\right)-x\left(-\frac{x}{y}\right)+y=0\ \ \ \text{[Using (i) and (ii)]}

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.