Simple one on first-order linear DEs

**emttim84** · April 23rd 2008, 11:30 AM

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.

**Mathstud28** · April 23rd 2008, 11:32 AM

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.

$6e^{x}(e^{x})=6(e^{x})^2=6e^{2x}$

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