Question

# Solve differential equationdy+5ydx=e^(-5x)dx

First order differential equations

Solve differential equation $$dy+5ydx=e^{-5x}dx$$

2021-01-14

$$\frac{dy}{dx}+P(x)y= Q(x)$$ (1)
where P(x) or Q(x) are constants or function of x alone
Integrating factor of (1) is
$$I.F.= e^{\int P(x)dx}$$
Required solution is
$$y (I.F.)= \int Q(x)(I.F)dx+C$$
$$dy+(5y)dx= e^{-5x}dx$$
$$\Rightarrow \frac{dy}{dx}+5y= e^{-5x} (*)$$
Since this equation is in the form $$\frac{dy}{dx}+P(x)y=Q(x)$$ so it is linear ODE comparing $$(*)$$ with $$\frac{dy}{dx}+P(x)y= Q(x)$$ we get $$P(x)=5$$, $$Q(x)= e^{−5x}$$
So integrating factor is
$$I.F.= e^{\int P(x)dx}$$
$$= e^{\int 5dx}$$
$$= e^{5x}$$
$$y(I.F.)= \int Q(x)(I.F.)dx+c$$
$$\Rightarrow y(e^{5x})= \int (e^{5x})(e^{5x})dx+C$$
$$\Rightarrow y(e^{5x})= \int dx+C$$
$$= y(e^{5x})= x+C$$
$$\Rightarrow y= \frac{x+C}{e^{5x}}$$