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Need solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 19 textbook solution.

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Answer : \text { (c) } 2 x+y^{3}-3 x^{2} y=0

Hint :

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

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

Multiply and divide by x^{2} in first term

\begin{aligned} &\Rightarrow x^{2} y^{3}\left(\frac{x d y-y d x}{x^{2}}\right)-x(x d y+y d x)=0 \\ &\Rightarrow x^{2} y^{3} d\left(\frac{y}{x}\right)-x d(x y)=0 \quad\left[d\left(\frac{y}{x}\right)=\frac{x d y-y d x}{x^{2}}, d(x y)=x d y+y d x\right] \end{aligned}

\Rightarrow x d(x y)-x^{2} y^{3} d\left(\frac{y}{x}\right)=0

Divide by x^{3}y^{2}

\Rightarrow \frac{d(x y)}{x^{2} y^{2}}=\frac{y}{x} d\left(\frac{y}{x}\right)

Integrate both sides

\begin{aligned} &\Rightarrow-\frac{1}{x y}=\frac{\left(\frac{y}{x}\right)^{2}}{2}+C \\ &\Rightarrow-\frac{1}{x y}-\frac{1}{2}\left(\frac{y}{x}\right)^{2}-C=0 \end{aligned}

\begin{aligned} &\Rightarrow-\left[\frac{1}{x y}+\frac{1}{2}\left(\frac{y}{x}\right)^{2}+C\right]=0 \\ &\Rightarrow \frac{1}{x y}+\frac{1}{2}\left(\frac{y}{x}\right)^{2}+C=0 \end{aligned}

Now, curve passes through (1,1)

\begin{aligned} &\Rightarrow \frac{1}{1}+\frac{1}{2}(1)+C=0 \\ &\Rightarrow \frac{2}{2}+\frac{1}{2}+C=0 \\ &\Rightarrow \frac{3}{2}+C=0 \end{aligned}

\begin{aligned} &\Rightarrow C=-\frac{3}{2} \\ &\Rightarrow \frac{1}{x y}+\frac{y^{2}}{2 x^{2}}-\frac{3}{2}=0 \\ &\Rightarrow \frac{2 x+y^{3}-3 x^{2} y}{2 x^{2} y}=0 \\ &\Rightarrow 2 x+y^{3}-3 x^{2} y=0 \end{aligned}



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