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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.


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)



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