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Please solve RD Sharma class 12 chapter Differential Equation exercise 21.2 question 16 subquestion (ix) maths textbook solution

Answers (1)

Answer:

 The required equation is

2xy\frac{\mathrm{d} y}{\mathrm{d} x}=(x^{2}+3y^{2})

Hint:

 Differentiating the given equation with respect to x

Given:

 x^{2}+y^{2}=ax^{3}

Solution:

\begin{aligned} &x^{2}+y^{2}=ax^{3} \\ &a=\frac{x^{2}+y^{2}}{x^{3}} \end{aligned}

Differentiating with respect to x,

\begin{aligned} &\frac{\left [ \left ( 2x+2y\frac{\mathrm{d} y}{\mathrm{d} x} \right )x^{3}-3x^{2}(x^{2}+y^{2}) \right ]}{x^{6}}=0 \\ &2x^{4}+2x^{3}y\frac{\mathrm{d} y}{\mathrm{d} x}-3x^{4}-3x^{2}y^{2}=0 \\ &2x^{3}y\frac{\mathrm{d} y}{\mathrm{d} x}-x^{4}-3x^{2}y^{2}=0 \end{aligned}

\begin{aligned} &2x^{3}y\frac{\mathrm{d} y}{\mathrm{d} x}=x^{4}+3x^{2}y^{2} \\ &2x^{3}y\frac{\mathrm{d} y}{\mathrm{d} x}=x^{2}(x^{2}+3y^{2}) \\ &2xy\frac{\mathrm{d} y}{\mathrm{d} x}=(x^{2}+3y^{2}) \end{aligned}

 The required equation is

2xy\frac{\mathrm{d} y}{\mathrm{d} x}=(x^{2}+3y^{2})

Posted by

Gurleen Kaur

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