# Getting into differential equations. I'm just getting into differential equations now and I've got to show that the given y(x) is a solution to the differential equation: u'+u=0, y(x)=Ce^{-x}.

Getting into differential equations
I'm just getting into differential equations now and I've got to show that the given y(x) is a solution to the differential equation:

How do I tackle this?
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Step 1
You know that the derivative of y(x)expx is a constant, so just compute it:
$0=\frac{d}{dx}\left[\mathrm{exp}\left(x\right)y\left(x\right)\right]=\mathrm{exp}\left(x\right)y\left(x\right)+\mathrm{exp}\left(x\right){y}^{\prime }\left(x\right)$
and as the exp is never $=0$, $y\left(x\right)+{y}^{\prime }\left(x\right)=0$.
Step 2
It also works the other way: if you consider a solution of the equation, then
$\frac{d}{dx}\left[\mathrm{exp}\left(x\right)y\left(x\right)\right]=0$
and then it has the given form.
###### Did you like this example?
Step 1
The point is that you plug $y\left(x\right)=C{e}^{-x}$ into the differential equation at the place of u , and of course the derivative of y(x) at the place of u′ (do you know how to differentiate y(x)?).
Step 2
Then you must show that the equality in the differential equation holds, that is ${y}^{\prime }\left(x\right)+y\left(x\right)=0$. So
$\begin{array}{r}{y}^{\prime }\left(x\right)+y\left(x\right)=-C{e}^{-x}+C{e}^{-x}=\left(-C+C\right){e}^{-x}=0\cdot {e}^{-x}=0.\end{array}$
So it is, in fact, a solution since it equals zero as it should.