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  • Provide solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.2 question 15 subquestion (i)

Provide solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.2 question 15 subquestion (i)

Answers (1)

Answer:

 2xy\frac{\mathrm{d} y}{\mathrm{d} x}+4x^{2}-y^{2}=0

Hint:

 Using chain rule and putting the value of a in (i)

Given:

 (2x+a)^{2}+y^{2}=a^{2} \qquad \qquad \dots (i)

Solution:

\begin{aligned} &2(2x+a)2+2y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &2(2x+a)+y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &(2x+a)=-\frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x}\\ &a=-\frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x}-2x \end{aligned}

\begin{aligned} &\left ( \frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y^{2}=\left ( \frac{y}{2}\frac{\mathrm{d} y}{\mathrm{d} x}+2x \right )^{2}\\ &\left ( \frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y^{2}=\left ( \frac{y}{2}\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+4x^{2}+2\frac{y}{2}.\frac{\mathrm{d} y}{\mathrm{d} x}.2x\\ &y^{2}-4x^{2}-2xy\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}

2xy\frac{\mathrm{d} y}{\mathrm{d} x}+4x^{2}-y^{2}=0

Posted by

Gurleen Kaur

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