# Simple equation re-arrangement

I have a simple re-arrangement of an equation which I can't seem to solve, help would be much appreciated.

I'm trying to re-arrange the equation:

$e^{-3t}\frac{dy}{dt} - 3e^{-3t}y = C$

where $C$ represents a constant.

Into:

$\frac{d}{dt}(e^{-3t}y) = C$

I have already tried multiple ways of doing this with no success, as I'm not too sure of what operations I can perform on C for it still to be classed as a constant (such as differentiating it).

You need to use the product formula for derivatives: $(f(t)g(t))'=f'(t)g(t)+f(t)g'(t)$.

In your case : $f(t)=e^{-3t}$ and $g(t)=y=y(t)$.

EDIT: Following our discussion in the comments, here is more details.

We compute the derivatives of $f$ and $g$ as above: $$f'(t)=\frac{d}{dt}(e^{-3t})=e^{-3t}\frac{d}{dt}(-3t)=e^{-3t}(-3)=-3e^{-3t}$$ and $$g'(t)=\frac{d}{dt}y$$ Altogether we get, $$\frac{d}{dt}(e^{-3t}y) = \frac{d}{dt}(f(t)g(t)) = f(t)\frac{d}{dt}g(t)+g(t)\frac{d}{dt}f(t)= e^{-3t}\frac{dy}{dt} - 3e^{-3t}y .$$ We conclude that $$C=e^{-3t}\frac{dy}{dt} - 3e^{-3t}y =\frac{d}{dt}(e^{-3t}y),$$ as desired.

• Thanks for the help, however I'm still not sure how to proceed from there. From using the product rule I've now got the same equation plus an extra (-3e^(-3t)y). I know there is something I'm missing but just can't figure it out. – Joe Apr 18 '15 at 10:06
• It must be in the way you compute the derivatives. It must be done like this : $\frac{d}{dt}(e^{-3t}y) =e^{-3t}\frac{dy}{dt} - 3e^{-3t}y$. – Thibaut Dumont Apr 18 '15 at 10:09
• That's exactly what I've done but that just results in: e^(-3t)dy/dt - 3e^(-3t)y -3e^(-3t)y = C. How would I get rid of both of the -3e^(-3t)y? – Joe Apr 18 '15 at 10:17
• I need clarification on what you want to do. Which equation do you start with? – Thibaut Dumont Apr 18 '15 at 10:17
• I started with dy/dt - 3y = Ce^(3t) and I'm trying to derive d/dt(e^(-3t)y) = C from it. – Joe Apr 18 '15 at 10:19

Set $z(t)=e^{-3t}y(t)$ and your equation reads $$z'(t)=C.$$

Integrating, you have

$$z(t)=Ct+C',$$ i.e. $$e^{-3t}y(t)=Ct+C',$$ thus $$y(t)=e^{3t}(Ct+C').$$