If $y=y(x)$ is a particular solution of $\sqrt{1-x^2} \frac{d y}{d x}+\frac{2 x}{\sqrt{1-x^2}}$ $y=x,…
If $y=y(x)$ is a particular solution of $\sqrt{1-x^2} \frac{d y}{d x}+\frac{2 x}{\sqrt{1-x^2}}$ $y=x, y(0)=1$, then $y\left(\frac{1}{2}\right)=$
- $\frac{\sqrt{3}}{2}$
- $\frac{1}{4}$
- $\frac{1}{2}$
- 0
Solution
$
\begin{aligned}
& \text {} \because \quad \sqrt{1-\mathrm{x}^2} \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2 \mathrm{x}}{\sqrt{1-\mathrm{x}^2}} \mathrm{y}=\mathrm{x} \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{y}=\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^2}}.....(i)
\end{aligned}
$
Which is a linear differential equation
$
\therefore \text { I.f. }=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{-\log \left(1-\mathrm{x}^2\right)} ; \text { I.F }=\frac{1}{1-\mathrm{x}^2}
$
Now, the solution to the differential equation (i) is
$
\begin{aligned}
& \text { y. I.F. }=\int \frac{x}{\sqrt{1-x^2}} \frac{1}{1-x^2} d x+c \\
& \Rightarrow y \times \frac{1}{1-x^2}=\frac{1}{-2} \int \frac{-2 x}{\left(1-x^2\right)^{\frac{3}{2}}} d x+c \\
& \Rightarrow \frac{y}{1-x^2}=-\frac{1}{2} \frac{\left(1-x^2\right)^{-\frac{1}{2}}}{\left(-\frac{1}{2}\right)}+c \\
& \Rightarrow y=\left(1-x^2\right)\left[\left(1-x^2\right)^{-1 / 2}+c\right].....(ii) \\
& \because y(0)=1, \text { then from eqn (ii), we get } \\
& y(0)=(1-0)\left[(1-0)^{-1 / 2}+c\right] \\
& \Rightarrow 1=1+c \Rightarrow c=0
\end{aligned}
$
Putting the value of $\mathrm{c}$ in $\mathrm{eq}^{\mathrm{n}}$ (ii), we get
$
\begin{aligned}
& y=\left(1-x^2\right)\left[\left(1-x^2\right)^{-1 / 2}+0\right] \\
& \Rightarrow y=\left(1-x^2\right)^{1 / 2} \\
& \text { At } x=\frac{1}{2} ; y\left(\frac{1}{2}\right)=\left(1-\frac{1}{4}\right)^{\frac{1}{2}}=\left(\frac{3}{4}\right)^{\frac{1}{2}} \\
& \Rightarrow y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}
\end{aligned}
$
Asked in: AP EAMCET 2023 (18 May Shift 2)
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