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

• luitzen
In summary, the conversation involves solving a differential equation and finding the particular solution. The equation can be rewritten as y' = t(5-2y) which is separable. If the equation is not separable, other methods such as multiplying by an integrating factor can be used. In this case, the integral e^t^2 is not necessary as it can be simplified to (ye^t^2)' = 5te^t^2. The conversation also touches on the use of integration by parts and finding the particular solution using u-substitution.
luitzen

## Homework Statement

Solve DE:y' + 2ty = 5t

## 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:
Hi luitzen!

Can't you just argue that

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

is equivalent with

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

and thus

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

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:)

luitzen said:
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?

Right! Seperable equations are easy!

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

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

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.

How would you integrate et2?

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

luitzen said:
How would you integrate et2?

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

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

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

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

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

LCKurtz said:
You wouldn't:
$$e^{t^2}y'+2te^{t^2}y=5te^{t^2}$$
$$(ye^{t^2})' =5te^{t^2}$$
If I follow that route, I'll get this:

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

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

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

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

No, no, no!

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

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.

## 1. What is a differential equation (DE)?

A differential equation is a mathematical equation that relates a function to its derivatives. It describes how a quantity changes over time or space, and is used to model many physical and natural phenomena.

## 2. What is the general solution to a differential equation?

The general solution to a differential equation is the most general form of the equation that satisfies all possible initial conditions. It contains one or more arbitrary constants and can be used to find specific solutions for different sets of initial conditions.

## 3. How do I solve a first-order linear differential equation?

To solve a first-order linear differential equation, you can use the method of integrating factors. This involves multiplying both sides of the equation by a suitable integrating factor, which can then be used to simplify the equation and solve for the unknown function.

## 4. What is the importance of solving differential equations in science?

Differential equations are used in many scientific fields to model and predict the behavior of systems, such as in physics, biology, economics, and engineering. Solving these equations allows us to understand and make predictions about the natural world and to develop new technologies.

## 5. What are some real-life applications of differential equations?

Some common applications of differential equations include modeling population growth, predicting the spread of diseases, analyzing chemical reactions, and understanding the behavior of electrical circuits. They are also used in fields such as fluid mechanics, celestial mechanics, and signal processing.

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