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

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

Answers (1)

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}

 

 

Posted by

infoexpert23

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