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 = e^2x.

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

y(I.F.) = \(\int\)(Q×I.F.)dx + C

⇒ye^2x = \(\int\)sin x . e^2x dx+C …....(1)

Let I = \(\int\)sin x . e^2x

⇒I = sin x . \(\int\)e^2xdx - \(\int\)(\(\frac{d}{dx}\)(sin x) . \(\int\)e^2xdx) 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.\(\int\)e^2x - \(\int\)(\(\frac {d}{dx}\)(cosx).∫e^2xdx)dx]

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

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

ye^2x = \(\frac {e^{2x}}{5}\)(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.