# Linear non-exact differential equation made exact

• rsera
In summary: Most of the tutors at my college are undergrad students, so they don't go that high in math. I asked one of them about this problem and he didn't know. And I don't think the other tutors ever took DE. You're right, I should have asked the professor for help. I'm just shy sometimes.
rsera

## Homework Statement

Find the general solution:
(y+1) dx + (4x - y) dy = 0

## Homework Equations

dy/dx + P(x)y = Q(x) (standard form)
e^(∫ P(x) dx) (integrating factor)

## The Attempt at a Solution

This exercise is in the chapter on linear equations, making non-exact equations exact.

So I know I need to put it into the form:
dy/dx + P(x)y = Q(x)

So that I can find the Integrating Factor:
e^( ∫ P(x) dx)

Then multiply through by the Integrating Factor and solve the resulting exact equation.

My problem with this one has been getting it into that standard form.

If I divide by dx and rearrange, I get:

(4x - y) dy/dx + (y+1) = 0

Then divide by (4x - y)

dy/dx + (y + 1) / (4x - y) = 0

However, this does not match the standard form, because it is not a function of x times y.

I know what to do once it's in standard form, I'm just having troubles getting to that point. Any clue as to what trick to use to get it to standard form would be appreciated. Thanks!

Last edited:
rsera said:

## Homework Statement

Find the general solution:
(y+1) dx + (4x - y) dy = 0

## Homework Equations

dy/dx + P(x)y = Q(x) (standard form)
e^(∫ P(x) dx) (integrating factor)

## The Attempt at a Solution

This exercise is in the chapter on linear equations, making non-exact equations exact.

So I know I need to put it into the form:
dy/dx + P(x)y = Q(x)

So that I can find the Integrating Factor:
e^( ∫ P(x) dx)

Then multiply through by the Integrating Factor and solve the resulting exact equation.

My problem with this one has been getting it into that standard form.

If I divide by dx and rearrange, I get:

(4x - y) dy/dx + (y+1) = 0

Then divide by (4x - y)

dy/dx + (y + 1) / (4x - y) = 0

However, this does not match the standard form, because it is not a function of x times y.

I know what to do once it's in standard form, I'm just having troubles getting to that point. Any clue as to what trick to use to get it to standard form would be appreciated. Thanks!
Hello rsera. Welcome to PF !

How about if you switch the roles of x & y? In other words, treat y as the independent variable and consider x to be a function of y .

1 person
Thank you for the welcome and the reply!

You know, I had thought about doing that after I posted, but wasn't sure if that was the right way to approach it.

So, if I switch them, I get the standard form as:
dx/dy + [4/(y+1)] * x = y / (y+1)

So my integrating factor is:
e^∫P(y) dy ; P(y) = 4 / (y+1)
e^ ∫4/(y+1) dy = e^ 4ln|y+1| = (y+1)^4

I can multiply through by (y+1)^4 and get:
(y+1)^4 * dx/dy + (y+1)^4 [4/(y+1)] * x = (y+1)^4 * (y / (y+1))

Which simplifies to:
(y+1)^4 * dx/dy + (y+1)^4 [4/(y+1)] * x = y(y+1)^3

And then integrate both sides to get:
(y+1)^4 * x = ∫y(y+1)^3 dy

Am I on the right track here? I tried integrating y(y+1)^3 using integration by parts, but I didn't get an answer that looked anything like the one in the book. If I'm correct up to here, can you suggest a better method for integrating it?

rsera said:
Thank you for the welcome and the reply!

You know, I had thought about doing that after I posted, but wasn't sure if that was the right way to approach it.

So, if I switch them, I get the standard form as:
dx/dy + [4/(y+1)] * x = y / (y+1)

So my integrating factor is:
e^∫P(y) dy ; P(y) = 4 / (y+1)
e^ ∫4/(y+1) dy = e^ 4ln|y+1| = (y+1)^4

I can multiply through by (y+1)^4 and get:
(y+1)^4 * dx/dy + (y+1)^4 [4/(y+1)] * x = (y+1)^4 * (y / (y+1))

Which simplifies to:
(y+1)^4 * dx/dy + (y+1)^4 [4/(y+1)] * x = y(y+1)^3

And then integrate both sides to get:
(y+1)^4 * x = ∫y(y+1)^3 dy

Am I on the right track here? I tried integrating y(y+1)^3 using integration by parts, but I didn't get an answer that looked anything like the one in the book. If I'm correct up to here, can you suggest a better method for integrating it?
That looks good.

What did you get for the integral?

1 person
The book has:
20x = 4y-1 + c(y+1)^-4

After working with it some more, I got:
20x = 5y-1 + c(y+1)^-4

For ∫y(y+1)^3 dy
I used u = y and dv = (y+1)^3; so, du = 1 dy and v = 1/4(y+1)^4
So I got:
∫u dv = y * 1/4 (y+1)^4 - 1/20 (y+1)^5 + c

Back to the DE, we have:
(y+1)^4 * x = (y)*(1/4)*(y+1)^4 - (1/20)*(y+1)^5 + c

I multiplied through by 20 to eliminate the fraction:
20*x*(y+1)^4 = 5*y*(y+1)^4 - (y+1)^5 + c

And factored out a (y+1)^4:
20*x*(y+1)^4 = (y+1)^4 (5y-1) +c

Finally, dividing by (y+1)^4 gives me:
20x = 5y-1 + c(y+1)^-4

So now my answer has a 5 where the answer in the book has a 4.

I really appreciate you looking over my work! It's nice to have someone to talk to about this. At my community college, the tutors don't go as high as DE, so getting help has been a real challenge.

rsera said:
The book has:
20x = 4y-1 + c(y+1)^-4

After working with it some more, I got:
20x = 5y-1 + c(y+1)^-4

For ∫y(y+1)^3 dy
I used u = y and dv = (y+1)^3; so, du = 1 dy and v = 1/4(y+1)^4
So I got:
∫u dv = y * 1/4 (y+1)^4 - 1/20 (y+1)^5 + c

Back to the DE, we have:
(y+1)^4 * x = (y)*(1/4)*(y+1)^4 - (1/20)*(y+1)^5 + c

I multiplied through by 20 to eliminate the fraction:
20*x*(y+1)^4 = 5*y*(y+1)^4 - (y+1)^5 + c

And factored out a (y+1)^4:
20*x*(y+1)^4 = (y+1)^4 (5y-1) +c
Check the algebra in that last step.
I get

20x(y+1)4 = (y+1)4 (5y - (y+1)) + c

Finally, dividing by (y+1)^4 gives me:
20x = 5y-1 + c(y+1)^-4

So now my answer has a 5 where the answer in the book has a 4.

I really appreciate you looking over my work! It's nice to have someone to talk to about this. At my community college, the tutors don't go as high as DE, so getting help has been a real challenge.

1 person
Yes! I fixed that and now my answer matches theirs! Thank you so much!

## 1. What is a linear non-exact differential equation?

A linear non-exact differential equation is a type of differential equation where the terms involving the dependent variable and its derivatives are linear, but the equation is not in an exact form. This means that it cannot be solved directly using standard methods and requires additional steps to make it exact.

## 2. Why is it important to make a non-exact differential equation exact?

Making a non-exact differential equation exact allows us to solve it using standard techniques and obtain an exact solution. This is important because it allows us to accurately model and predict the behavior of systems described by the equation.

## 3. How can a linear non-exact differential equation be made exact?

A linear non-exact differential equation can be made exact by finding an integrating factor, which is a function that multiplies the entire equation to make it exact. This can be done by using various methods such as the method of integrating factors or the method of exact equations.

## 4. Are there any limitations or conditions for making a non-exact differential equation exact?

Yes, there are certain conditions that must be met in order to make a non-exact differential equation exact. These conditions include the equation being linear, having a specific form, and having a certain type of solution. Not all non-exact differential equations can be made exact.

## 5. Can a non-exact differential equation be solved without making it exact?

In some cases, a non-exact differential equation can be solved without making it exact. This can be done by using alternative methods such as separation of variables or using a computer to approximate a solution. However, making the equation exact is often the preferred method as it leads to an exact solution.

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