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 $\displaystyle x^2$ or $\displaystyle 2x$ in this case so bear with me.

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

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

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

Then I did $\displaystyle e^x(y'+y)=(6e^x)e^x$ and that's where I'm forgetting my exponent rules...would $\displaystyle (6e^x)e^x$ become $\displaystyle 6e^{x^2}$ or $\displaystyle 6e^{2x}$?
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 $\displaystyle x^2$ or $\displaystyle 2x$ in this case so bear with me.

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

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

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

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