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

Answer: (C)

$\frac{1}{6}$

$\frac{dy}{dx}+\frac{2xy}{(1+x^2)}=\frac{4x^2}{(1+x^2)}$

This differential equation is of the form $\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$

Solution of above differential equation is $y{e}^{\int p\left(x\right)dx}=\int e^{\int p\left(x\right)dx}q\left(x\right)dx+c$

In this problem, $p\left(x\right)=\frac{2x}{1+x^2}$ and $q\left(x\right)=\frac{4x^2}{(1+x^2)}$

$\therefore \int p\left(x\right)dx=\int \frac{2x}{1+x^2}dx=ln(1+x^2)$

$⟹e^{\int p\left(x\right)dx}=e^{ln(1+x^2)}=1+x^2$

$\int e^{\int p\left(x\right)dx}q\left(x\right)dx$

$=\int (1+x^2)\left(\frac{4x^2}{1+x^2}\right)dx$

$=\int 4x^{2}dx$

$=\frac{4x^3}{3}$

So, the solution of the differential equation is

$y{e}^{\int p\left(x\right)dx}=\int e^{\int p\left(x\right)dx}q\left(x\right)dx+c$

$⟹y(1+x^2)=\frac{4x^3}{3}+c$

$⟹y=\frac{4x^3}{3(1+x^2)}+\frac{c}{1+x^2}$

Using the condition $y\left(0\right)=-1$ we get

$y\left(0\right)=c=-1$

So, the exact solution is

$⟹y=\frac{4x^3}{3(1+x^2)}-\frac{1}{1+x^2}$

$⟹y\left(1\right)=\frac{4}{3(1+1)}-\frac{1}{1+1}$

$=\frac{1}{6}$

So, the answer is option (C).