Question

Solve the differential equation $(1-x^2)\frac{dy}{dx}+xy=a.$

• A. $-y=-ax+c\sqrt{(1-x^2)}.$
• B. $y=-ax+c\sqrt{(1-x^2)}.$
• C. $-y=ax+c\sqrt{(1-x^2)}.$
• D. None of these

Answer 1

Answer: (D)

None of these

$(1+x^2)=\frac{dy}{dx}+xy=a$$\Rightarrow \frac{dy}{dx}+\frac{x}{(1-x^2)}y=\frac{a}{(1-x^2)}$ ...(1)
Here $P=\frac{x}{(1+x^2)}$ $\Rightarrow \int Pdx=\int \frac{x}{(1+x^2)}dx.$

$=-\frac{1}{2}\log (1-x^2)=\log \frac{1}{\sqrt{1-x^2}}$
$\therefore I.F=e^{\log \frac{1}{\sqrt{1+x^2}}}=\frac{1}{\sqrt{1-x^2}}$
Multiplying (1) by $I.F$ we get
$\frac{1}{\sqrt{1-x^2}}\frac{dy}{dx}+\frac{x}{{(1-x^2)}^{\frac{3}{2}}}y=\frac{a}{{(1-x^2)}^{\frac{3}{2}}}$
Integrating both sides
$\frac{y}{\sqrt{1-x^2}}=\int \frac{a}{{(1-x^2)}^{\frac{3}{2}}}dx+c$
Put $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$
$\therefore \frac{y}{\sqrt{1-x^2}}=\int a{\sec }^2\theta dx+c=a\tan \theta +c=\frac{ax}{\sqrt{1-x^2}}+c$
$\therefore y=ax+c\sqrt{1-x^2}$