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

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

Answers (1)

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,

y=vx

\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}

Posted by

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