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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 48 maths textbook solution.

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Answer : c=4 x^{3} y-2 y^{4}

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

Given:  \left(x^{3}-2 y^{3}\right) d x+3 x^{2} y d y=0

Solution : \left(x^{3}-2 y^{3}\right) d x+3 x^{2} y d y=0

\begin{aligned} &3 x^{2} y d y+\left(x^{3}-2 y^{3}\right) d x=0 \\ &3 x^{2} y d y=\left(2 y^{3}-x^{3}\right) d x \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{2 y^{3}-x^{3}}{3 x^{2} y} \\ &\frac{d x}{d y}=\frac{-3 x^{2} y}{x^{3}-2 y^{3}} \\ &=\frac{3}{\frac{-x}{y}+\frac{2 y^{2}}{x^{2}}} \\ &=\frac{3}{2\left(\frac{y}{x}\right)^{2}-\frac{x}{y}} \end{aligned}

Put \frac{y}{x}=y, differentiate both sides,


\begin{aligned} &\frac{d y}{d x}=v+\frac{x d v}{d x} \\ &v+\frac{x d v}{d x}=\frac{3}{2 v^{2}-\frac{1}{v}} \\ &\frac{x d v}{d x}=\frac{3 v-2 v^{4}+v}{2 v^{3}-1} \end{aligned}

Integrating both sides,

\begin{aligned} &\int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{2 \mathrm{v}^{3}-1}{3 \mathrm{v}-2 \mathrm{v}^{4}+\mathrm{v}} \mathrm{d} \mathrm{V} \\ &\int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{2 \mathrm{v}^{3}-1}{4 \mathrm{v}-2 \mathrm{v}^{4}} \mathrm{~d} \mathrm{~V} \end{aligned}

put, 4 v-2 v^{4}=t

\begin{aligned} &\left(4-8 v^{3}\right) d v=d t \\ &-4\left(2 v^{3}-1\right) d v=d t \\ &\int \frac{d x}{x}=\frac{-1}{4} \int \frac{d t}{t} \\ &\log x+\log c=\frac{-1}{4} \log t \end{aligned}

\begin{aligned} &\log (c x)=\log t^{\frac{-1}{4}} \\ &t^{\frac{-1}{4}}=c x \end{aligned}

Put value of t

\left(4 v-2 v^{4}\right)^{\frac{-1}{4}}=c x

put value of v=\frac{y}{x}

\begin{aligned} &\Rightarrow\left(\frac{4 y}{x}-\frac{2 y^{4}}{x^{4}}\right)^{\frac{-1}{4}}=\mathrm{cx} \\ &\Rightarrow\left(\frac{4 x^{3} y-2 y^{4}}{x^{4}}\right)^{\frac{-1}{4}}=c x \\ &\Rightarrow \frac{x}{\left(4 x^{3} y-2 y^{4}\right)^{\frac{1}{4}}}=c x \quad\left[\because\left(x^{4}\right)^{\frac{-1}{4}}=x^{-1}\right] \end{aligned}

\begin{aligned} &\Rightarrow \frac{1}{\left(4 \mathrm{x}^{3} \mathrm{y}-2 \mathrm{y}^{4}\right)^{\frac{1}{4}}}=\mathrm{c} \\ &\Rightarrow \frac{1}{\mathrm{c}}=\left(4 \mathrm{x}^{3} \mathrm{y}-2 \mathrm{y}^{4}\right)^{\frac{1}{4}} \\ &\Rightarrow \sqrt[4]{\mathrm{c}}=4 \mathrm{x}^{3} \mathrm{y}-2 \mathrm{y}^{4} \\ &\Rightarrow \quad \mathrm{c}=4 \mathrm{x}^{3} \mathrm{y}-2 \mathrm{y}^{4} \end{aligned}

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