Finding the integral in y' + (2/t)y = cos(t)/t²

[Bashyboy]

Homework Statement

y' + (2/t)y = (cost)/(t^2), and the following condition is given: y(pi) = 0

Homework Equations

The Attempt at a Solution

After employing the integrating factor, I find the solution to be:

$y=e^{-2t} \int e^{2t} \frac{\cos(t)}{t^2} dt$.

Evidently, this simplifies all the way to y = (sin t)/(t^2). I am not sure as to how this integral should be solved. Any hints would be much welcomed.

 

[Curious3141]

I think you should show all your steps, starting with exactly how you applied your integrating factor. Clearly, there's an error somewhere.

 

[Bashyboy]

I see what I did wrong.

mu of t, the integrating factor, $\mu (t) = e^{\int \frac{2}{t}}dt = e^{2 \ln t} = e^{2t}$

Do you see where I went wrong? It should be $\mu (t) = 2t$

 

[HallsofIvy]

This is an example of what happens when you memorize formulas (imperfectly) rather than learning basic definitions and how the formulas are derived.

An "integrating factor" is a function, $\mu(t)$ such that multiplying the equation by it converts the left side into a single derivative. Here, that means we must have $\mu y'+ (2\mu/t)y= (\mu y)'$. Expanding the derivative on the right that becomes $\mu y'+ \mu' y= \mu y'+ (2\mu/t)y$ which reduces to $\mu'= 2\mu/t$, a separable differential equation for $\mu$. $d\mu/\mu= 2dt/t$ integrates to $ln(\mu)= 2ln(t)$ or $\mu= t^2$ NOT "2t" (I have neglected the "constant of integration since we only need a single function).

Multiplying the entire equation by $t^2$ gives $t^2y'+ 2ty= (t^2y)'= cos(t)$ which is easy to integrate.

 

[Curious3141]

I see what I did wrong.

mu of t, the integrating factor, $\mu (t) = e^{\int \frac{2}{t}}dt = e^{2 \ln t} = e^{2t}$

Do you see where I went wrong? It should be $\mu (t) = 2t$

$e^{2 \ln t} = (e^{\ln t})^2 = t^2$, that's your integrating factor.

Your integrating factor is NOT $2t$. $2t$ is in fact the derivative of your integrating factor, and you should be able to see this from applying product rule to $yt^2$.