Question

$\frac{dy}{dx}+x\sin 2y=x^3{\cos }^{2}y.$
Find the particular curve which passes through $(0,\pi /4).$What is the value of 2 times the coefficient of $e^{-x^2}$

Answer 1

$\frac{dy}{dx}+x\sin 2y=x^3{\cos }^{2}y\Rightarrow {\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^3$

Put $\tan y=v\Rightarrow {\sec }^{2}y.\frac{dy}{dx}=\frac{dv}{dx}$

$\therefore \frac{dv}{dx}+2xv=x^3$ ....(1)

Here $P=2x\Rightarrow \int Pdp=\int 2xdx=x^2$

$\therefore I.F.=e^{x^2}$

Multiplying (1) by $I.F.$ we get

$e^{x^2}\frac{dv}{dx}+e^{x^2}.2xv=e^{x^2}.x^3$

Integrating both sides, we get

$v.e^{x^2}=\int x^3.e^{x^2}dx+c$

Put $x^2=t\Rightarrow 2xdx=dt$

$\therefore v.e^t=\frac{1}{2}\int t{e}^{t}dt=\frac{1}{2}{e}^t(t-1)+c$

$\Rightarrow \tan y=\frac{1}{2}(x^2-1)+c{e}^{-x^2}$

If this curve passes through $(0,\frac{\pi }{4})$, then $1=-\frac{1}{2}+c\Rightarrow c=\frac{3}{2}$

$\therefore \tan y=\frac{1}{2}(x^2-1)+\frac{3}{2}{e}^{-x^2}$