Question

Find the general solution: $\frac {dy}{dx}+2y=sin\ x$

Answer 1

The given differential equation is $\frac {dy}{dx}$+2y = sin x.

This is in the form of $\frac {dy}{dx}$+py = Q (where p=2 and Q=sin x).
$\int$pdx = $\int$2dx = 2$\int$1dx = 2x

Now, I.F = e∫pdx = e2x.

The solution of the given differential equation is given by the relation,

y(I.F.) = $\int$(Q×I.F.)dx + C

⇒ye2x = $\int$sin x . e2x dx+C …....(1)

Let I = $\int$sin x . e2x

⇒I = sin x . $\int$e2xdx - $\int$($\frac{d}{dx}$(sin x) . $\int$e2xdx) dx

⇒I = sin x . $\frac {e^{2x}}{2}$ - $\int$(cos x . $\frac {e^{2x}}{2}$)dx

⇒I = $\frac {e^{2x}sin\ x}{2}$ - $\frac 12$[cos x.$\int$e2x - $\int$($\frac {d}{dx}$(cosx).∫e2xdx)dx]

⇒I = $\frac {e^{2x}sin\ x}{2}$ - $\frac 12$[cos x . e2x/2-$\int$[(-sinx).e2x/2]dx]

⇒I = $\frac {e^{2x}sin\ x}{2}$ -$\frac {e^{2x}cos\ x}{4}$-\frac 14$$\int(sin x . e2x)dx

⇒I = $\frac {e^{2x}}{4}$(2sin x - cos x)-$\frac 14$I

⇒$\frac 54$I = $\frac {e^{2x}}{4}$(2sin x - cos x)

⇒I = $\frac {e^{2x}}{5}$(2sin x - cos x)

Therefore, equation(1)becomes:

ye2x = $\frac {e^{2x}}{5}$(2sin x - cos x)+C

⇒y = $\frac 15$(2sin x -cos x) + Ce-2x

This is the required general solution of the given differential equation.