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

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

Answers (1)

Answer:

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}

 

 

Posted by

infoexpert21

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