First slide

Methods of solving first order first degree differential equations

Question

$If x^{3} \frac{d y}{d x} = y^{3} + y^{2} \sqrt{y^{2} - x^{2}} and y = 1 when x = 1 then y = 2 when x is$

Difficult

A

2/5
B

4/5
C

6/5
D

8/5

Solution

$\frac{d y}{d x} = \frac{y^{3}}{x^{3}} + \frac{y^{2}}{x^{3}} x \sqrt{\frac{y^{2}}{x^{2}} - 1}$
$\begin{array}{l} Put y = Vx \to \frac{dy}{dx} = V + x \frac{dv}{dx} \\ \to V + x \frac{dv}{dx} = v^{3} + v^{2} \sqrt{v^{2} - 1} \\ x \frac{d v}{d x} = v^{3} - v + v^{2} \sqrt{v^{2} - 1} \\ x \frac{dv}{dx} = v (v^{2} - 1) + v^{2} \sqrt{v^{2} - 1} \\ x \frac{d v}{d x} = \sqrt{v^{2} - 1} (v \sqrt{v^{2} - 1} + v^{2}) \\ \frac{d v}{\sqrt{v^{2} - 1} (v \sqrt{v^{2} - 1} + v^{2})} = \frac{d x}{x} \\ \to \int \frac{dx}{x} = \int \frac{dv}{v \sqrt{v^{2} - 1} (v + \sqrt{v^{2} - 1})} \times \frac{(v - \sqrt{v^{2} - 1})}{(v - \sqrt{v^{2} - 1})} \\ \int \frac{dx}{x} = \int \frac{(v - \sqrt{v^{2} - 1})}{v \sqrt{v^{2} - 1}} d v \\ \int \frac{dx}{x} = \int \frac{1}{\sqrt{v^{2} - 1}} d v - \int \frac{1}{v} dv \\ \log x + \log c = \log (v + \sqrt{v^{2} - 1}) - \log v \\ \Rightarrow c x = \frac{v + \sqrt{v^{2} - 1}}{v} = \frac{y + \sqrt{y^{2} - x^{2}}}{y} \\ Put x = y = 1 \Rightarrow c = 1 \\ \Rightarrow p u t y = 2 \Rightarrow x = \frac{2 + \sqrt{4 - x^{2}}}{2} \\ 2 x = 2 + \sqrt{4 - x^{2}} \Rightarrow {(2 x - 2)}^{2} = 4 - x^{2} \\ \Rightarrow (5 x - 8) x = 0 \to x = \frac{8}{5} \end{array}$

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