Question

The solution of the differential equation $\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{{\text{sin}}^{2}x}{1+x^3}$ is

• A. $y(1+x^3)=x+\frac{1}{2}\text{sin}2x+c$
• B. $y(1+x^3)=cx+\frac{1}{2}\text{sin}2x$
• C. $y(1+x^3)=cx-\frac{1}{2}\text{sin}2x$
• D. $y(1+x^3)=\frac{x}{2}-\frac{1}{4}\text{sin}2x+c$

Answer 1

The correct option is D

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

$\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{{\text{sin}}^{2}x}{1+x^3}$

Here, $P=\frac{3x^2}{1+x^3}\Rightarrow \text{I.F}=e^{\int Pdx}=e^{\text{log}(1+x^3)}=1+x^3$

Thus the solution is

$y.(1+x^3)=\int \frac{{\text{sin}}^{2}x}{1+x^3}(1+x^3)dx=\int \frac{1-\text{cos}2x}{2}dx$

$\Rightarrow y(1+x^3)=\frac{1}{2}x-\frac{\text{sin}2x}{4}+\text{c}$