Solution of the differential equation $y^{\prime}=\frac{\left(x^2+y^2\right)}{x y}$, where $y(1)=-2$ is…
Solution of the differential equation $y^{\prime}=\frac{\left(x^2+y^2\right)}{x y}$, where $y(1)=-2$ is given by
- $y^2=4 x^2 \log x^2+x^2$
- $y^2=x^2 \log x-x^2$
- $y^2=x \log x^2+4 x^2$
- $y^2=x^2 \log x^2+4 x^2$
Solution
$\begin{aligned} & \frac{d y}{d x}=\frac{x^2+y^2}{x y}=\frac{x}{y}+\frac{y}{x} \\ & \text { Let } \frac{x}{y}=v \Rightarrow y=\frac{v}{x} \\ & \therefore \frac{d y}{d x}=\frac{v-x\left(\frac{d v}{d x}\right)}{v^2} \\ & \therefore \frac{v-x\left(\frac{d v}{d x}\right)}{v^2}=v+\frac{1}{v} \Rightarrow \frac{1}{v}-\left(\frac{x}{v^2}\right) \frac{d v}{d x}=v+\frac{1}{v} \\ & \therefore \int \frac{d v}{v^3}=-\int \frac{d x}{x} \\ & \therefore \frac{-1}{2 v^2}=-\log x-c \quad \Rightarrow \frac{1}{2 v^2}=\log x+c \\ & \therefore \frac{y^2}{2 x^2}=\log \mathrm{x}+\mathrm{c} \quad \Rightarrow \mathrm{y}^2=\mathrm{x}^2 \log \mathrm{x}^2+2 \mathrm{x}^2 \mathrm{c} \\ & \text { We have } \mathrm{y}(1)=-2 \\ & \therefore 4=0+2 c \Rightarrow c=2 \\ & \therefore \mathrm{y}^2=\mathrm{x}^2 \log \mathrm{x}^2+4 \mathrm{x}^2 \\ & \end{aligned}$
Asked in: MHT CET 2021 (22 Sep Shift 1)
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