Question

For the differential equation, find the general solution:

`dy/dx  + 2y = sin x`

Answer 1

The given equation is `dy/dx + 2y = sin x.`            ....(1)
Which is a linear equation of type `dy/dx + Py = Q.`

Here P = 2 and Q = sin x.

∴ `I.F. = e^(int Pdx) = e^(int2 dx) = e^(2x)`

∴ The solution is `y .(I.F.) =int Q. (I. F.) dx + C`

`y. e^(2x) = int e^(2x) sin x  dx + C = I + C`         ....(2)

Now, `I = int e^(2x) sin x  dx`

`= e^(2x) (- cos x) - int 2e^(2x) (- cos x)  dx`   ...[Integrating by part] 

`= -e^(2x) cos x + 2 int e^(2x)  cosx  dx`

Again integrating parts,

`I = - e^(2x) cos x + 2 [e^(2x) sin x - int e^(2x) * 2 sin x  dx]`

`= I = - e^(2x) cos x + 2e^(2x) sin x - 4I`

⇒ 5I = e^2x (2 sin x - cos x)

⇒`I = (e^(2x))/5 (2 sin x - cos x)`            ....(3)

Substituting the value of (3) in (2), we get

`y. e^(2x) = 1/5 e^(2x) (2 sin x - cos x) + C`

⇒ `y = 1/5 (2 sin x - cos x) + Ce^(-2x),`

Which is the required solution.