Question

Solve `dy/dx+x sin 2 y=x^3 cos^2 y` 

 

Answer 1

Given, `dy/dx+x sin 2 y=x^3 cos^2 y` 

Dividing both sides by cos²x,

`Sec^2 x dy/dx+x sec^2x sin 2y=x^3` 

`Sec^2 x dy/dx+2x tan y=x^3....................(1)` 

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

`Sec^2 x dy/dx=(dv)/dx`  

Hence, from (1), we get `(dv)/dx+2v.x=x^3` 

∴ `P=2x and Q=x^3` 

∴` int p  dx=int 2x  dx =x^2` 

∴ `I.F.= e^int "^(p  dx) = e^int 2x  dx=e^(x^2)` 

∴ The solution is v ` e^(x^2)=int e^(x^2) x^3 dx+c` 

To find the integral put `x^2=t, xdx=dt/2` 

`∴I= inte^t.t. dt/2=1/2[te^t-int e^t.dt]..................` [By  parts]

∴` I= 1/2 [te^t-e^t]=1/2e^t(t-1)=1/2e^"x^2 (x^2-1)`  

∴ The solution is v `e^x^2=1/2 e^x"^2(x^2-1)+c` 

∴ tan `y  e x^2= 1/2 e^x"^2 (x^2-1)+c`