# Thread: Finding a Solution using an integrating constant

1. ## Finding a Solution using an integrating constant

Err meant integrating factor not constant. My book is terrible and is confusing the heck out of me. The equation is y'-2y=3e^t and I'm supposed to find a general solution and draw a direction field as well as describe how solutions behave for large t. For the solution I've taken the integrating factor m to be e^-2t and the equation to be my'-2my=m3e^t

That yields y'e^-2t-2ye^-2t=e^-2t 3e^t
or y'e^-2t-2ye^-2t=3e^-t

I'm stumped at this point. Can someone help me out?

2. Originally Posted by sfgiants13
Err meant integrating factor not constant. My book is terrible and is confusing the heck out of me. The equation is y'-2y=3e^t and I'm supposed to find a general solution and draw a direction field as well as describe how solutions behave for large t. For the solution I've taken the integrating factor m to be e^-2t and the equation to be my'-2my=m3e^t

That yields y'e^-2t-2ye^-2t=e^-2t 3e^t Mr F says: The whole point of getting the integrating factor is to write the left hand side of this equation as ${\color{red} \frac{d}{dt} [y e^{-2t}]}$ ....

or y'e^-2t-2ye^-2t=3e^-t

I'm stumped at this point. Can someone help me out?

3. Originally Posted by sfgiants13
Err meant integrating factor not constant. My book is terrible and is confusing the heck out of me. The equation is y'-2y=3e^t and I'm supposed to find a general solution and draw a direction field as well as describe how solutions behave for large t. For the solution I've taken the integrating factor m to be e^-2t and the equation to be my'-2my=m3e^t

That yields y'e^-2t-2ye^-2t=e^-2t 3e^t
or y'e^-2t-2ye^-2t=3e^-t

I'm stumped at this point. Can someone help me out?
Multiply $y'-2y=3e^t$ on $e^{-2t}$, then

$e^{-2t}y'-2e^{-2t}y=3e^{-t}$

$e^{-2t}y'+\left(e^{-2t}\right)'y=3e^{-t}$

$\left(e^{-2t}y\right)'=3e^{-t}$

You know what to do now?

4. In general, $\mu(x)$ is an "integrating factor" for the differential equation $\frac{dy}{dx}+ p(x,y)= q(x)$ if multiplying by it makes the left side of the equation an "exact differential". That is, if $\mu(x)\frac{dy}{dx}+ \mu(x)p(x,y)= \frac{d\mu(x)y}{dx}$. By the product rule, of course, $\frac{d\mu(x)y}{dx}= \mu(x)\frac{dy}{dx}+ \frac{d\mu}{dx}y$ so that means we must have $\frac{d\mu}{dx}= \mu p(x,y)$, a differential equation for $\mu$. Theoretically, every differential equation has an integrating factor but for general p(x,y) that equation can be as hard to solve as the original equation.

However, for linear equations, where p(x,y) is just p(x) times y, the differential equation for $\mu(x)$, $\frac{d\mu}{dx}= \mu(x)p(x)$, is "separable" and easy to solve. "Separate" as $\frac{d\mu}{\mu}= p(x)dx$ and integrate: $ln(\mu(x))= \int p(x)dx$ so $\mu(x)= e^{\int p(x)dx}$.

For this particular equation, $\frac{dy}{dt}-2y=3e^t$, p(t)= -2 so $\int p(t)dt= -2t$ (since you only need an integral, you can ignore the constant of integration) and the integrating factor is, just as you say, $e^{-2t}$.

Multiplying the equation by that gives $e^{-2t}\frac{dy}{dt}- 2e^{-2t}y= 3e^te^{-2t}= 3e^{-t}$.

That's exactly where you got to. Now, the whole point of an "integrating factor" is that $\frac{de^{-2t}y}{dt}= e^{-2t}\frac{dy}{dt}- 2e^{-2t}y$! You can differentiate using the product rule to check but that is, as I say, the whole point of the "integrating factor".

Now, your differential equation is $\frac{de^{-2t}y}{dt}= 3e^{-t}$. Just integrate. The integral on the left, by the "fundamental theorem of calculus", is just $e^{-2t}y$, of course.