Solve differential equation dy+5ydx=e^(-5x)dx

Question
Solve differential equation $$dy+5ydx=e^(-5x)dx$$

2021-01-14
$$dy/dx+P(x)y= Q(x)$$ (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
$$y (I.F.)= int Q(x)(I.F)dx+C$$
$$dy+(5y)dx= e^(-5x)dx$$
$$=> dy/dx+5y= e^(-5x)\( (*) Since this equation is in the form \(dydx+P(x)y=Q(x)$$ so it is linear ODE comparing (*) with $$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$$
$$=> y(e^5x)= int (e^5x)(e^5x)dx+C$$
$$=> y(e^5x)= int dx+C$$
$$= y(e^5x)= x+C$$
$$=> y= (x+C)/e^(5x)$$

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