Question

Let $y\left(x\right)$ be a solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ and $y\left(0\right)=–1$. Then $y\left(1\right)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $-1$

Answer 1

The correct option is C $\frac{1}{6}$

$(1+x^2)\frac{dy}{dx}+2xy-4x^2=0\Rightarrow \frac{dy}{dx}+\left(\frac{2x}{1+x^2}\right)y=\frac{4x^2}{1+x^2}$
$I.F.=e^{\int \frac{2x}{1+x^2}dx}=e^{\log (1+x^2)}=1+x^2$
$\Rightarrow y(1+x^2)=\int 4x^{2}dx+c\Rightarrow y(1+x^2)=\frac{4x^3}{3}+c$
$y\left(0\right)=–1\Rightarrow -1=c$
$y(1+x^2)=\frac{4x^3}{3}-1\Rightarrow y=\frac{\frac{4x^3}{3}-1}{(1+x^2)}$
$\Rightarrow y\left(1\right)=\frac{1}{6}$