The general solution of the differential equation $\left(1-x^{2}\right) \frac{d y}{d x}+2 x…

The general solution of the differential equation $\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}}$ is
  1. $y=\sqrt{1-x^{2}}+c\left(1-x^{2}\right)$
  2. $y=2 \sqrt{1-x^{2}}+c$
  3. $y=2 \sqrt{1-x^{2}}+c\left(1+x^{2}\right)$
  4. $y \sqrt{1-x^{2}}=c\left(1-x^{2}\right)$

Solution

$\begin{aligned} &\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}} \\ \therefore & \frac{d y}{d x}+\frac{2 x y}{1-x^{2}}=\frac{x}{\left(1-x^{2}\right)^{\frac{1}{2}}} \\ \text { I.F. }=& e^{\int \frac{2 x}{1-x^{2}} d x}=e^{-\int \frac{-2 x}{1-x^{2}} d x}=e^{-\log \left(1-x^{2}\right)}=e^{\log \left(\frac{1}{1-x^{2}}\right)}=\frac{1}{1-x^{2}} \end{aligned}$ $\begin{aligned} \therefore\left(\frac{1}{1-x^{2}}\right) &=\int \frac{x}{\left(1-x^{2}\right)^{\frac{1}{2}}} \times \frac{1}{\left(1-x^{2}\right)} d x \\ &=\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)^{3 / 2}} d x=\left(-\frac{1}{2}\right) \frac{\left(1-x^{2}\right)^{-\frac{1}{2}}}{\left(-\frac{1}{2}\right)}+c \\ y\left(\frac{1}{1-x^{2}}\right) &=\frac{1}{\sqrt{1-x^{2}}+c \Rightarrow y}=\sqrt{1-x^{2}}+c\left(1-x^{2}\right) \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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