Solution of the first order equation $y'=\lambda y-\lambda e^{\lambda t}$

Question

I've been trying to solve this first order ODE $y'=\lambda y-\lambda e^{\lambda t}$ but to no avail.

Here is what I've done.

$y'=\lambda( y- e^{\lambda t})$

$\implies \int\frac{1}{y-e^{\lambda t}}\frac{dy}{dt}dt=\int\lambda dt$

$\implies \int\frac{1}{y-e^{\lambda t}}dy=\int\lambda dt$

$\implies \ln \left( y-{{\rm e}^{\lambda\,t}} \right)=\lambda t+c$

$\implies y-{{\rm e}^{\lambda\,t}}=Ae^{\lambda t}$ where $A=e^c.$

$\implies y=(A+1)e^{\lambda t}$ but Maple is providing the solution as $y=(A-t\lambda)e^{\lambda t}$. Where I'm I missing it?

Answer 1

Hint: Your equation has the form $y'+p(t)y=q(t)$. This is a first order, non-homogeneous differential equation. This type of differential equation can be solved by applying variation of parameters. Rearranging your equation: $y'-\lambda y=-\lambda e^{\lambda t}$

In your case $p(t)=-\lambda$ and $q(t)=-\lambda e^{\lambda t}$.

Answer 2

$y'=\lambda y-\lambda e^{\lambda t}$

$-\lambda e^{\lambda t}= y'-\lambda y =e^{\lambda t}\left(e^{-\lambda t}y(t)\right)' $

$\left(e^{-\lambda t}y(t)\right)' = -\lambda $

$e^{-\lambda t}y(t) = -\lambda t +c$

$y(t) = -\lambda te^{\lambda t} +ce^{\lambda t},$ $c\in \mathbb R$

Answer 3

As Kenny mentioned in the comments, you cannot treat t as a constant in the integral on the left side of your equation while you integrate t as a variable on the right. Also as projectilemotion mentioned in the comments, this problem can be solved via integrating factor.

$y' = \lambda y - \lambda e^{\lambda t}$

$\implies y' - \lambda y = - \lambda e^{\lambda t}$

Now multiply by the integrating factor $e^{-\lambda t}$ to get

$y'e^{- \lambda t} - \lambda y e^{-\lambda t} = - \lambda$

$\implies (y e^{- \lambda t})' = - \lambda$

Then integrate with respect to t to get

$ y e^{- \lambda t} = - \lambda t + A$

So that

$y = -\lambda t e^{\lambda t} + A e^{\lambda t} = (A - \lambda t)e^{\lambda t}$