Question

For the differential equation: $\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{{\sin }^{2}x}{1+x^3}$. The solution is $y(1+x^3)=\frac{1}{A}x-\frac{1}{B}\sin 2x+c$, where c is the constant of integration, then find B/A?

Answer 1

$\frac{dy}{dx}+\frac{{3x}^2}{1+x^3}y=\frac{{\sin }^{2}x}{1+x^3}$ ...(1)
Here $P=\frac{{3x}^2}{1+x^3}\Rightarrow \int Pdx=\int \frac{{3x}^2}{1+x^3}dx=\log (1+x^3)$
$\therefore I.F=e^{\log (1+x^3)}=1+x^3$
Multiplying (1) by $I.F,$ we get
$(1+x^3)\frac{dy}{dx}+{3x}^{2}y={\sin }^{2}x$
Integrating both sides
$(1+x^3)y=\int {\sin }^{2}xdx+c=\frac{1}{2}x-\frac{1}{4}\sin 2x+c$
$\therefore \frac{B}{A}=\frac{4}{2}=2$