# Solve Diff. Eq.: Dy/Dx + 1 + y + x=(x+y)^2 e^(3x)

• Jtechguy21
In summary: Good job. In summary, after utilizing the substitution u=x+y and simplifying the equation, the next step is to get rid of all the y's in the ODE, including dy/dx. Then, another substitution can be used to simplify the equation further. Continuing with the process of solving for p, the integrating factor can be determined and used to solve for p. Finally, by substituting back in the original variable u=x+y, the solution for y can be determined.
Jtechguy21

## Homework Statement

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

Since u=x+y
Du/dx=1dy/dx

## 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 isn't in the form

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

maybe I am approaching this problem incorrectly?
any guidance would be appreciated

Jtechguy21 said:

## Homework Statement

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

Since u=x+y
Du/dx=1dy/dx

## 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 isn't in the form

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

maybe I am approaching this problem incorrectly?
any guidance would be appreciated

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.

Dick said:
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.

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?

Jtechguy21 said:
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?

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.

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.

Jtechguy21 said:
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.

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

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

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?

Jtechguy21 said:
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?

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

thanks for walking me through this problem

Jtechguy21 said:

thanks for walking me through this problem

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

1 person
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:
Jtechguy21 said:
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 :)

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.

## 1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model various physical, biological, and social phenomena.

## 2. How do you solve a differential equation?

To solve a differential equation, one must find the function that satisfies the equation. This can be done using various methods such as separation of variables, substitution, or using specific techniques for different types of differential equations.

## 3. What is the order of a differential equation?

The order of a differential equation is the highest derivative present in the equation. For example, a first-order differential equation contains only the first derivative, while a second-order equation contains the second derivative.

## 4. What is the particular solution of a differential equation?

The particular solution of a differential equation is a specific solution that satisfies the equation under given initial conditions. It is different from the general solution, which includes all possible solutions to the equation.

## 5. How is a differential equation used in science?

Differential equations are used in many scientific fields, including physics, engineering, biology, and economics, to model and understand complex systems. They are also used to make predictions and solve real-world problems.

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