Question

Find $y$, if

$(1-x^2)\frac{dy}{dx}+2xy=x(1-x^2{)}^{1/2}$.

• A. $y=c(1+y^2)+\sqrt{1-x^2}$
• B. $x=c(1-y^2)+\sqrt{1-y^2}$
• C. $y=c(1-x^2)+\sqrt{1-y^2}$
• D. $y=c(1-x^2)+\sqrt{1-x^2}$

Answer 1

Answer: (D)

$y=c(1-x^2)+\sqrt{1-x^2}$

$(1-x^2)\frac{dy}{dx}+2xy=x(1-x^2{)}^{\frac{1}{2}}$
On rearranging terms, we have,
$\frac{dy}{dx}+\frac{2xy}{(1-x^2)}=\frac{x}{(1-x^2{)}^{\frac{1}{2}}}$
This is in the form of $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ can be solved using integration factor $\left(IF\right)$ method
$IF=e^{\int P\left(x\right)dx}=e^{\int \frac{2x}{(1-x^2)}dx}$
Let $(1-x^2)=t\Rightarrow -2xdx=dt$.
Consider $IF=e^{\int \frac{2x}{(1-x^2)}dx}=e^{\int \frac{2x}{(1-x^2)}dx}=e^{\int \frac{-dt}{t}}=\frac{1}{t}$
Solution of the equation: $y\cdot \left(IF\right)=\int \frac{x}{(1-x^2{)}^{\frac{1}{2}}}\cdot \left(IF\right)dx+c$
$\frac{y}{(1-x^2)}=\int \frac{xdx}{(1-x^2{)}^{\frac{3}{2}}}+c$
Consider $\int \frac{xdx}{(1-x^2{)}^{\frac{3}{2}}}$$=\int \frac{-dt}{2(t{)}^{\frac{3}{2}}}=\frac{1}{(t{)}^{\frac{1}{2}}}=\frac{1}{(1-x^2{)}^{\frac{1}{2}}}$
$\therefore y=c(1-x^2)+(1-x^2{)}^{\frac{1}{2}}$