Question

If $y\left(x\right)$ satisfies equations $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ and $y\left(0\right)=0$, then $y\left(1\right)$ is

• A. $\frac{2}{3}$
• B. $1$
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{2}{3}$

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