# Thread: Simple one on first-order linear DEs

1. ## Simple one on first-order linear DEs

Alright, I know this is a simple exponent issue, but I can't remember if it's $x^2$ or $2x$ in this case so bear with me.

The problem states to find the particular solution that satisfies the initial condition.

$y'+y=6e^x$ is the DE and the conditions are $y=3$ when $x=0$

$P(x)=1$ in this case so the $IF=e^{\int1dx}=e^x$

Then I did $e^x(y'+y)=(6e^x)e^x$ and that's where I'm forgetting my exponent rules...would $(6e^x)e^x$ become $6e^{x^2}$ or $6e^{2x}$?
Sorry for the stupid question.

2. Originally Posted by emttim84
Alright, I know this is a simple exponent issue, but I can't remember if it's $x^2$ or $2x$ in this case so bear with me.

The problem states to find the particular solution that satisfies the initial condition.

$y'+y=6e^x$ is the DE and the conditions are $y=3$ when $x=0$

$P(x)=1$ in this case so the $IF=e^{\int1dx}=e^x$

Then I did $e^x(y'+y)=(6e^x)e^x$ and that's where I'm forgetting my exponent rules...would $(6e^x)e^x$ become $6e^{x^2}$ or $6e^{2x}$?
Sorry for the stupid question.
$6e^{x}(e^{x})=6(e^{x})^2=6e^{2x}$