Question

Solve ${\displaystyle \frac{dy}{dx}+x\sin 2y=x^3{\cos }^{2}y}$ 

 

Answer 1

Given, ${\displaystyle \frac{dy}{dx}+x\sin 2y=x^3{\cos }^{2}y}$ 

Dividing both sides by cos²x,

${\displaystyle {\mathrm{Sec}}^{2}x\frac{dy}{dx}+x{\sec }^{2}x\sin 2y=x^3}$ 

${\displaystyle {\mathrm{Sec}}^{2}x\frac{dy}{dx}+2x\tan y=x^3....................\left(1\right)}$ 

Put tan y = v and differentiate w.r.t. x, 

${\displaystyle {\mathrm{Sec}}^{2}x\frac{dy}{dx}=\frac{dv}{dx}}$  

Hence, from (1), we get ${\displaystyle \frac{dv}{dx}+2v.x=x^3}$ 

∴ ${\displaystyle P=2x\text{and}Q=x^3}$ 

∴${\displaystyle \int p dx=\int 2x dx=x^2}$ 

∴ ${\displaystyle I.F.=e^{\int }{}^{p dx}=e^{\int }2x dx=e^{x^2}}$ 

∴ The solution is v ${\displaystyle e^{x^2}=\int e^{x^2}{x}^{3}dx+c}$ 

To find the integral put ${\displaystyle x^2=t,xdx=\frac{dt}{2}}$ 

${\displaystyle \therefore I=\int e^t.t.\frac{dt}{2}=\frac{1}{2}[t{e}^t-\int e^t.dt]..................}$ [By  parts]

∴${\displaystyle I=\frac{1}{2}[t{e}^t-e^t]=\frac{1}{2}{e}^t(t-1)=\frac{1}{2}{e}^{}{x}^2(x^2-1)}$  

∴ The solution is v ${\displaystyle e^x^2=\frac{1}{2}{e}^x{}^2(x^2-1)+c}$ 

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