The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d}…
The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y+\sqrt{x^2-y^2}}{x}$ is
- $\sin ^{-1} y=\log x+\mathrm{c}$, where c is a constant of integration.
- $\frac{y}{x}=\sin ^{-1} x+\mathrm{c}$, where c is a constant of integration.'
- $\frac{y}{x}=\sqrt{x^2-y^2}+\mathrm{c}$, where c is a constant of integration.
- $\sin ^{-1}\left(\frac{y}{x}\right)=\log x+\mathrm{c}$, where c is a constant of integration.
Solution
$\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y+\sqrt{x^2-y^2}}{x} ...(i)\\
& \text { Put } y=\mathrm{v} x ...(ii)\\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{~d} v}{\mathrm{~d} x}
...(iii)\end{aligned}$
Substituting (ii) and (iii) in (i), we get
$\begin{array}{ll} & \mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x}=\frac{\mathrm{v} x+\sqrt{x^2-\mathrm{v}^2 x^2}}{x} \\ \therefore & \mathrm{v}+x \frac{\mathrm{~d} v}{\mathrm{~d} x}=\mathrm{v}+\sqrt{1-\mathrm{v}^2} \\ \therefore & x \frac{\mathrm{dv}}{\mathrm{d} x}=\sqrt{1-\mathrm{v}^2} \\ \therefore & \int \frac{\mathrm{dv}}{\sqrt{1-\mathrm{v}^2}}=\int \frac{\mathrm{d} x}{x} \\ \therefore & \sin ^{-1}(\mathrm{v})=\log |x|+\mathrm{c} \\ \therefore & \sin ^{-1}\left(\frac{y}{x}\right)=\log |x|+\mathrm{c}\end{array}$
Asked in: MHT CET 2024 (09 May Shift 2)
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