# Homework Help: DE:y' + 2ty = 5t

1. Jun 24, 2011

### luitzen

1. The problem statement, all variables and given/known data
Solve DE:y' + 2ty = 5t

3. The attempt at a solution
For yh: y' + 2ty = 0
y' = -2ty
Thus int(y-1,y) = int(-2t,t)
ln(y) = -t2 + C1
y = e-t2 + C1 = eC1e-t2 = Ce-t2

To find yp there are two ways to do it. With an integrating factor exp(-A(t)) where A'(t) = 2t or by saying yp = v(t)yh(t). Both ways will result in int(e-t2,t) which is unsolvable.

So I checked up the answer which read 5/2 + Cet2, so yp(t)=5/2. So I checked it by entering 5/2 in the DE and of course it was right.

Now my question is, how would you arrive at that answer? When would you make the assumption yp' = 0 and check if it is right? Isn't that too trivial? And when you would find an answer for yp' = 0, would you stop searching for another answer?

Last edited: Jun 24, 2011
2. Jun 24, 2011

### micromass

Hi luitzen!

Can't you just argue that

$$y^\prime+2ty=5t$$

is equivalent with

$$y^\prime=t(5-2y)$$

and thus

$$\frac{y^\prime}{5-2y}=t$$

3. Jun 24, 2011

### luitzen

Yes, of course. The question was: Check if the following equations are separable equations and if yes, give the solution.

So, if you have a separable equation, you don't need to find yh and yp, am I right?

If that's not the case, you try to find yh and after that yp, right?

Thank you very much:)

4. Jun 24, 2011

### micromass

Right! Seperable equations are easy!

Indeed, if the equation is not seperable, then you'll need to do other tricks. For example, working with yh and yp...

5. Jun 24, 2011

### LCKurtz

In my opinion, the best way to have worked that problem is to observe that it is a linear first order so you can multiply by $e^{t^2}$ making an exact derivative.

6. Jun 24, 2011

### luitzen

How would you integrate et2?

7. Jun 24, 2011

### vela

Staff Emeritus
You shouldn't have to. How are you coming up with that integral?

8. Jun 24, 2011

### LCKurtz

You wouldn't:
$$e^{t^2}y'+2te^{t^2}y=5te^{t^2}$$
$$(ye^{t^2})' =5te^{t^2}$$

9. Jun 25, 2011

### luitzen

Ok this is what I did:
$2ty-5t+y'=0$

$e^{t^{2}}\left(2ty-5t\right)+e^{t^{2}}y'=0$

$\dfrac{\partial F\left(y,t\right)}{\partial t}=e^{t^{2}}\left(2ty-5t\right)$

$\dfrac{\partial F\left(y,t\right)}{\partial y}=e^{t^{2}}$

10. Jun 25, 2011

### luitzen

If I follow that route, I'll get this:

$\left(ye^{t^{2}}\right)'=5te^{t^{2}}$

$ye^{t^{2}}=\int5te^{t^{2}}dt$

$\int5te^{t^{2}}dt=5t\int e^{t^{2}}dt-5\int\int e^{t^{2}}dtdt$

$y=5e^{t^{-2}}\left(t\int e^{t^{2}}dt-\int\int e^{t^{2}}dtdt\right)$

11. Jun 25, 2011

### HallsofIvy

No, no, no!

Let $u= t^2$. Then du= 2t dt, tdt= du/2.

12. Jun 25, 2011

### luitzen

lol, I think I use integration by parts too quickly.

It's not too hard, but I still don't get it why it's easier though.