The solution of the differential equation d y d x = - ( x 2 + 3 y 2 3 x 2 + y 2 ) , y ( 1 ) = 0 is [2023]

Question

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}), y (1) = 0$ is [2023]

Smart Achievers · Accepted Answer

(4)

Given, $\frac{d y}{d x} = - [\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}], y (1) = 0$

$Substitute y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$v + x \frac{d v}{d x} = - [\frac{x^{2} + 3 v^{2} x^{2}}{3 x^{2} + v^{2} x^{2}}] = - [\frac{1 + 3 v^{2}}{3 + v^{2}}]$

$\Rightarrow x \frac{d v}{d x} = - [\frac{1 + 3 v^{2}}{3 + v^{2}}] - v$ $= - [\frac{1 + 3 v^{2} + 3 v + v^{3}}{3 + v^{2}}]$

$\Rightarrow x \frac{d v}{d x} = - \frac{{(1 + v)}^{3}}{3 + v^{2}} \Rightarrow \int \frac{3 + v^{2}}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$\Rightarrow \int \frac{1 + v^{2} + 2 v - 2 v + 2}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$\Rightarrow \int \frac{1}{(1 + v)} d v + 2 \int \frac{1 - v}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$Substitute 1 + v = t \Rightarrow v = t - 1 \Rightarrow d v = d t$

$∴$ $\int \frac{d t}{t} + 2 \int \frac{1 - t + 1}{t^{3}} d t = - \ln x + C$

$\Rightarrow \log_{e} t + 2 \int (\frac{2 - t}{t^{3}}) d t = - \log_{e} x + C$

$\Rightarrow \log_{e} t + 2 \int (2 t^{- 3} - t^{- 2}) d t = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| \frac{y}{x} + 1 |$ $+ 2 [- t^{- 2} + t^{- 1}] = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| \frac{y + x}{x} |$ $+ 2$ $[\frac{- 1}{{(1 + \frac{y}{x})}^{2}} + \frac{1}{\frac{y}{x} + 1}] = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| y + x |$ $+ 2$ $[\frac{- x^{2}}{{(y + x)}^{2}} + \frac{x}{y + x}]$ $= C$

$\Rightarrow \log_{e}$ $| y + x |$ $+ \frac{2 x y}{{(y + x)}^{2}} = C$

$We have y (1) = 0 \Rightarrow C = 0 .$

$\Rightarrow \log_{e}$ $| x + y |$ $+ \frac{2 x y}{{(x + y)}^{2}} = 0$

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