# Diff. Eq. Solve

1. Mar 3, 2014

### Jtechguy21

1. The problem statement, all variables and given/known data

Solve
Dy/Dx + 1 + y + x=(x+y)^2 e^(3x)
Hint: Use the substitution u=x+y

2. Relevant equations
Since u=x+y
Du/dx=1dy/dx

3. The attempt at a solution

So basically the first thing I did was utilize my substitution.
Dy/Dx + 1 +y + x=u^(2)e^(3x)

next i solved for y. y=u-x

Dy/Dx +1+(u-x)+x=U^2*e^(3x)
My x's cancel out

Dy/Dx +1+u=u^2*e^(3x)

I divided both sides by u^2

Dy/dx+(1/u^2)+(1/u)=e^(3x)

I have no idea what to do next.
since this isnt in the form

Dy/dx+p(x)y=f(x) <-Linear form

maybe im approaching this problem incorrectly?
any guidance would be appreciated

2. Mar 3, 2014

### Dick

Start with a correction. If u=x+y then du/dx=1+dy/dx. The next step is to get rid of all of the y's in the ODE. Including dy/dx. Do that and then maybe think about another substitution.

3. Mar 3, 2014

### Jtechguy21

using
u=x+y
du/dx=1+dy/dx
and
y=u-x

My equation now becomes
Du/dx + u-x+x=u^(2)e^(3x)
Simplify (eliminated x's and divided by u^2)

Du/dx 1/u=e^(3x)
p(x)=1 e^∫1dx= e^x as my integrating factor

e^x du/dx +e^x(1/u)=e^4x

d/dx[e^x (1/u)]=e^4x now i take the integral of e^4x= (1/4)e^4x

e^x(1/u)=(1/4)e^4x+c

divide by e^x

1/u=(1/4)e^3x+ce^-x

plug back in u

1/(x+y)=(1/4)e^3x+ce^-x

Does this look better?

4. Mar 3, 2014

### Dick

I get to du/dx+u=u^2*e^(3x). Then your solution gets murky. Dividing by u^2 gives me (du/dx)/u^2+1/u=e^(3x). That seemed like a good time for another substitution.

5. Mar 3, 2014

### Jtechguy21

du/dx+u=u^2*e^(3x).
oh i see what your saying. i assumed when i divide by u^2 the du/dx didnt get it affected.

6. Mar 3, 2014

### Dick

It sure does get affected. See if you can think of a clever substitution to continue after the division by u^2.

Last edited: Mar 3, 2014
7. Mar 3, 2014

### Jtechguy21

I have no idea where to use a substitution
but after dividing by u^2 I get

(1/u^2)du/dx +1/u=e^3x

p=1/u
Dp/Dx=-1/u^2 du/dx
=========^^^^^^^^^^This looks like the left side of my equation.

(i pulled out the negative as a scalar so it matches my equation)

-dp/dx + p =e^3x how does that look so far?

8. Mar 3, 2014

### Dick

Looks great! Keep going! That's a pretty simple form.

9. Mar 3, 2014

### Jtechguy21

thanks for walking me through this problem

10. Mar 3, 2014

### Dick

Very welcome, but you've got a sign wrong. Write that as dp/dx-p=(-e^(3x)). P(x)=(-1)!

11. Mar 3, 2014

### Jtechguy21

oh i see what you did there. you multiplied by -1.
ill make the correction right now.

Dp/dx - p=-e^(3x) p(x)=(-1)

That means my integrating factor is
e^-(x)

e^-(x)dp/dx -e^-(x)p=-e^(2x)

d/dx[e^-(x)P] = -e^2x
take the integral of ^
-(1/2)e^2x

e^-(x)P= -(1/2)e^2x + c
p=1/u
divide the e^-(x)

1/u= -(1/2)e^3x+ce^x

u=x+y

1/x+y= -(1/2)e^3x+ce^x

how does that look?

how would i solve for Y? :)

that took a long time to figure this whole thing out, but its satisfying feeling.
thanks once again dick for helping me, and baby stepping me along the way lol :)

Last edited: Mar 3, 2014
12. Mar 3, 2014

### Dick

That looks much better. You don't necessarily have to solve for y. But I think if you put your mind to it, you can. Just try. You aren't that bad at this.