Question

y^{'}+6y=4e^{-5x}

Answer 1

y^{'}+6y=4e^{-5x}

The problem is rewritten as

$\frac{dy}{dx} + 6y = 4e^{-5x}$

This is a first order linear differential equation

Integrating factor = $e^{\smallint6dx} = e^{6x}$

Multiplying both sides by the integrating factor we get

$e^{6x}[\frac{dy}{dx} + 6y] = 4e^{-5x}e^{6x}$

=> $e^{6x}\frac{dy}{dx} + 6e^{6x}y = 4e^{x}$

=> $e^{6x}\frac{dy}{dx} + \frac{de^{6x}}{dx}y = 4e^{x}$

=> $\frac{d(ye^{6x})}{dx} = 4e^{x}$

=> $d(ye^{6x})= 4e^{x}dx$

Integrating both sides

$\int{d(ye^{6x})}=\int{ 4e^{x}dx}$

=> $ye^{6x} = 4e^{x}+C$ where C is integration constant