Question

Using integrating factor, solve the differential equation frac {dy}{dx}+y=e^{x}

Answer 1

The linear first order differential equation:

$\frac {dy}{dx} + P(x)y = Q(x)$

has the integrating factor (IF) : $IF= e^{\int P (x) dx}$ . 

(1) $\frac {dy}{dx}+y=e^{x}$ .

Integrating factor:

$P(x) = 1$ .

Integrating factor, $IF= \int e^{\int P (x) dx} = e^{\int dx}= e^{x}$

$IF= e^{x}$

Multiply equation by IF:

$e^x\frac {dy}{dx} + e^xy = e^xe^x$

So,

$e^x\frac {dy}{dx} + e^xy = e^{2x}$ ,

i.e. $\frac {d}{dx}[ e^xy] = e^{2x}$

Integrate:

$$e^xy = \int e^{2x} dx$$

i.e.

$$e^xy = \frac {1}{2}e^{2x} + C ,$$

$$y = e^{-x}[\frac {1}{2}e^{2x} + C].$$

.

Therefore, general solution of a Differential Equation (1):

$$y = \frac {1}{2}e^{x} + Ce^{-x}$$

.