Question

Find the solution of

{(y \frac{d y}{d x} + 2 x)}^{2} = (y^{2} + 2 x^{2}) [1 + {(\frac{d y}{d x})}^{2}]

and hence find the curve passing through

(1, 0) .

Accepted Answer

${(y \frac{d y}{d x} + 2 x)}^{2} = (y^{2} + 2 x^{2}) [1 + {(\frac{d y}{d x})}^{2}]$

$\Rightarrow {(\frac{d y}{d x})}^{2} - 2 \frac{y}{x} \frac{d y}{d x} + \frac{1}{2} {(\frac{y}{x})}^{2} - 1 = 0$

$\Rightarrow \frac{d y}{d x} = \frac{2 \frac{y}{x} \pm \sqrt{4 \frac{y^{2}}{x^{2}} - 4 (\frac{1}{2} \frac{y^{2}}{x^{2}} - 1)}}{2}$

$\Rightarrow \frac{d y}{d x} = \frac{y}{x} \pm \sqrt{\frac{y^{2}}{2 x^{2}} + 1} \dots (i)$

This is homogeneous differential equation

Let, $\frac{y}{x} = v \Rightarrow y = v x$

$\Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \dots (ii)$

Value from $(ii)$ putting in $(i)$ , we get

$v + x \frac{d v}{d x} = v \pm \sqrt{\frac{v^{2}}{2} + 1}$

$\Rightarrow \frac{d v}{\sqrt{\frac{v^{2}}{2} + 1}} = \pm \frac{d x}{x} \Rightarrow \int \sqrt{2} \frac{d v}{\sqrt{v^{2} + 2}} = \pm \int \frac{d x}{x}$

$\Rightarrow \sqrt{2} \cdot \log |v + \sqrt{v^{2} + 2}| = \pm \log x + \log c$

Replacing $v,$ we get

$\Rightarrow \sqrt{2} \log (\frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}} + 2}) = \pm \log x + \log c$

$∵$ Curve passes through $(1, 0)$

$\Rightarrow \sqrt{2} \log \sqrt{2} = \log c$

Therefore solution

$\sqrt{2} \log |\frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}} + 2}| = \pm \log x + \sqrt{2} \log \sqrt{2}$

$\Rightarrow \log (\frac{y}{x} + \sqrt{\frac{y^{2} + 2 x^{2}}{x^{2}}}) = \log \sqrt{2} \pm \frac{1}{\sqrt{2}} \log x$

K. C. Sinha Solutions for Chapter: Differential Equations - II, Exercise 2: EXERCISE

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha Solutions for Chapter: Differential Equations - II, Exercise 2: EXERCISE

Questions from K. C. Sinha Solutions for Chapter: Differential Equations - II, Exercise 2: EXERCISE with Hints & Solutions

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