Originally Posted by

**sirellwood** sorry i am new to this site so havent mastered LaTex yet but question is fairly straightforward to write....

The acceleration, x'' , of an object which is moving in a straight line, is

-2x'+2+e^(-t)

where x(t) is the distance travelled (and t is time). Write down the

differential equation for x(t). Find its general solution and then that

solution which satisfies x(0) = 0, x'(0) = 0;

Hi **sirellwood**.

You have

$\displaystyle x''\ =\ -2x'+2+e^{-t}$

Integrate with respect to $\displaystyle t.$

$\displaystyle x'\ =\ -2x+2t-e^{-t}+C$

where $\displaystyle C$ is a constant. That is your differential equation for $\displaystyle x(t)$ (which you can rearrange as $\displaystyle x'+2x=2t-e^{-t}+C).$ You can also find $\displaystyle C$ immediately using the given fact that $\displaystyle x(0)=x'(0)=0.$