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  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 38 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 38 Maths Textbook Solution.

Answers (1)

Answer: 2A=log\left ( xy \right )-\frac{x}{y}

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

Given:\frac{dy}{dx}=\frac{y\left ( x-y \right )}{x\left ( x+y \right )}

Solution:\frac{dy}{dx}=\frac{y\left ( x-y \right )}{x\left ( x+y \right )}

x\left ( x+y \right )dy=y\left ( x-y \right )dx    .....................(1)

Put ,  y=vx  and differentiate both side.

\frac{dy}{dx}=x\frac{dv}{dx}+v

Eq. (1) becomes,

\begin{aligned} &x(x+y x)(x d x+y d x)=y x(x-y x) d x \\ &x^{2}(1+v)(x d v+y d x)-v x^{2}(1-v) d x=0 \\ &x^{2}[(1+v) x d v+v(1+v) d x-v(1-v) d x]=0 \\ &x(1+v) d v+v(1+v) d x-v(1-v) d x=0 \\ &{[y(1+v) d x-v(1-v) d x]=-x(1+v) d v} \\ &{\left[v+v^{2}-v+v^{2}\right] d x=-x(1+v) d v} \\ &2 v^{2} d x=-x(1+v) d v \\ &\frac{d x}{-x}=\frac{1}{2 v^{2}}+\frac{v}{2 v^{2}} \int d v \\ &\frac{d x}{-x}=\frac{1}{2} v^{-2}+\frac{11}{2 v} \int d v \end{aligned}

Now, integrating both sides,

\begin{aligned} &\int \frac{\mathrm{dx}}{-\mathrm{x}}=\frac{1}{2} \int \mathrm{v}^{-2} \mathrm{dv}+\frac{1}{2} \int \frac{1}{\mathrm{v}} \mathrm{dv} \\ &-\log |\mathrm{x}|+\mathrm{A}=\frac{1}{2} \frac{\mathrm{v}^{-2+1}}{-2+1}+\frac{1}{2} \log |\mathrm{v}| \\ &-\log |\mathrm{x}|+\mathrm{A}=-\frac{1}{2} \mathrm{v}^{-1}+\frac{1}{2} \log \mathrm{v} \\ &-\log |\mathrm{x}|+\mathrm{A}=\frac{-1}{2 \mathrm{v}}+\log \mathrm{v}^{\frac{1}{2}} \\ &\mathrm{~A}=\frac{-1}{2 \mathrm{v}}+\log \mathrm{v}^{\frac{1}{2}}+\log |\mathrm{x}| \\ &\mathrm{A}=\frac{-1}{2 \mathrm{v}}+\log \mathrm{v}_{2} x \end{aligned}

Put value of  v=\frac{y}{x}

\begin{aligned} &\mathrm{A}=\frac{-1}{2\left(\frac{y}{\mathrm{x}}\right)}+\log (\mathrm{xy})^{\frac{1}{2}} \\ &\mathrm{~A}=\frac{-\mathrm{x}}{2 \mathrm{y}}+\frac{1}{2} \log \mathrm{xy} \\ &2 \mathrm{~A}=\log (x y)-\frac{\mathrm{x}}{\mathrm{y}} \end{aligned}                                                        (where A is integration constant)

 

 

Posted by

infoexpert21

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