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Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.3 question 16 maths

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y=c e^{\tan ^{-1} x}  is a solution of differential equation


Differentiate with respect to x

Use formula \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}


y=c e^{\tan ^{-1} x}


Differentiating on both sides with respect to x

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(c e^{\tan ^{-1} x}\right) \\\\ &\frac{d y}{d x}=c e^{\tan ^{-1} x}\left(\frac{1}{1+x^{2}}\right) \end{aligned}                                ................(i)

Differentiating equation (i) with respect to x

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=c \frac{d}{d x}\left(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\right) \end{aligned}

\frac{d^{2} y}{d x^{2}}=c\left[\frac{\left(1+x^{2}\right) \frac{d}{d x}\left(e^{\tan ^{-1} x}\right)-e^{\tan ^{-1} x} \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right]

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=c\left[\frac{\left(1+x^{2}\right) \frac{e^{\tan ^{-1} x}}{1+x^{2}}-e^{\tan ^{-1} x}(2 x)}{\left(1+x^{2}\right)^{2}}\right] \\\\ &\frac{d^{2} y}{d x^{2}}=c\left[\frac{e^{\tan ^{-1} x}-2 x e^{\tan ^{-1} x}}{\left(1+x^{2}\right)^{2}}\right] \end{aligned}

\frac{d^{2} y}{d x^{2}}=\left[\frac{c(1-2 x) e^{\tan ^{-1} x}}{\left(1+x^{2}\right)^{2}}\right]                                .............(ii)

Put (i) and (ii) in given equation

\begin{aligned} &\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0 \\\\ &L H S=\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x} \end{aligned}

\begin{aligned} &=\left(1+x^{2}\right) \frac{c(1-2 x) e^{\tan ^{-1} x}}{\left(1+x^{2}\right)^{2}}+(2 x-1) \frac{c e^{\tan ^{-1} x}}{\left(1+x^{2}\right)} \\\\ &=\frac{c(1-2 x) e^{\tan ^{-1} x}}{\left(1+x^{2}\right)}-(1-2 x) \frac{c e^{\tan ^{-1} x}}{\left(1+x^{2}\right)} \\\\ &=0 \\\\ &=R H S \end{aligned}

Thus, y=c e^{\tan ^{-1} x} is a solution of the differential equation.


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