Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 37  Maths Textbook Solution.

$y-2x=cx^{2}y$

Hint: you must know the rules of solving differential equation and integrations.

Given:  $\frac{dy}{dx}+\frac{y}{x}=\frac{y^{2}}{x^{2}}$

Solution: $\frac{dy}{dx}+\frac{y}{x}=\frac{y^{2}}{x^{2}}$

$\frac{dy}{dx}=\frac{y^{2}}{x^{2}}-\frac{y}{x}$

$\frac{dy}{dx}=\frac{y^{2}xy}{x^{2}}$

Put y = v x and differentiate both sides w.r.t  x

\begin{aligned} &\frac{d y}{d x}=v .1+x \frac{d v}{d x} \\ &\frac{d y}{d x}=v+x \frac{d v}{d x} \end{aligned}

So, equation (I) becomes

\begin{aligned} &\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}^{2}-\mathrm{v} \\ &\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}^{2}-\mathrm{v}-\mathrm{v} \\ &\mathrm{x} \frac{\mathrm{dv}}{\mathrm{d} \mathrm{x}}=\mathrm{v}^{2}-2 \mathrm{v} \\ &\left(\frac{1}{\mathrm{v}^{2} 2 \mathrm{v}}\right) \mathrm{d} \mathrm{v}=\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}

Now, integrating both sides

\begin{aligned} &\int\left(\frac{1}{(v-1)^{2}-1} d v=\int \frac{1}{x} d x\left[\because(v-1)^{2}-1=v^{2}+1-2 v-1=v^{2}-2 v\right]\right. \\ &\frac{1}{2} \log \left(\frac{v-2}{v}\right)=\log |x|+\log c \\ &\frac{1}{2} \log \left(\frac{v-2}{v}\right)=\log c x \\ &\log \left(\frac{v-2}{v}\right)=2 \log c x \\ &\log \left(\frac{v-2}{v}\right)=\log c^{2} x^{2}\left[\because \operatorname{alog} x=\log x^{a}\right] \end{aligned}

\begin{aligned} &\frac{\mathrm{v}-2}{\mathrm{v}}=\mathrm{c}^{2} \mathrm{x}^{2}\\ &1-\frac{2}{v}=c x^{2}[c=c]\\ &\left.1-\frac{2}{\frac{y}{x}}=c x^{2} \text { (put value of } v=\frac{y}{x}\right)\\ &1-\frac{2 x}{y}=c x^{2}\\ &y-2 x=c x^{2} y \end{aligned}

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