How Do You Rearrange Complex Differential Equations into Standard Linear Form?

[ch2kb0x]

Homework Statement

(2e^y -x) dy/dx = 1

Homework Equations

dy/dx + P(x)y = Q(x)

The Attempt at a Solution

I know how to solve these equations, but I can't get this into the dy/dx + P(x)y = Q(x) form.

In addition to this problem, is also this: (x + y^2)dy = ydx (can't get into the form). Help would be appreciated.

Also, for both of these questions, they are asking to consider x as function of x = x(y).

 

[LCKurtz]

Homework Statement

(2e^y -x) dy/dx = 1


Also, for both of these questions, they are asking to consider x as function of x = x(y).

Use the hint, using dy/dx = 1 / (dx/dy) you can rewrite it as:

dx/dy + x = 2e^y

It looks strange because you are used to thinking of y as a function of x. If the x and y were switched it would look familiar:

dy/dx + y = 2e^x

You could even switch them and work it on a sheet of scratch paper, then copy it back with the x's and y's switched back if that helps you visualize it.

 

[ch2kb0x]

So, if I were to solve, would it be like this:

dx/dy + x = 2e^y

I(x) = e^integral dy = e^y

multiply integrating factor on both sides will give: e^y (x) = e^(2y) + C

=> x = e^y + Ce^-y.

That was the answer I got for x, but it does not match answer in the back.

help.

 

[LCKurtz]

So, if I were to solve, would it be like this:

dx/dy + x = 2e^y

I(x) = e^integral dy = e^y

multiply integrating factor on both sides will give: e^y (x) = e^(2y) + C

=> x = e^y + Ce^-y.

That was the answer I got for x, but it does not match answer in the back.

help.

It's hard to get used to switching the independent and dependent variables y and x. Your integrating factor isn't I(x) now, it's I(y). Your solution looks OK other than that notational thing. And, regardless what the solution manual may say, you can check that your version of the solution solves the above equation.

 

[ch2kb0x]

Hmmm ok, still kind of confused. this is what I did after your correction:

dx/dy + x = 2e^y

I(y) = e^integral (2e^y) dy = e^(2e^y)

multiply integrating factor on both sides will give: e^(2e^y) x = Integral e^(2e^y) (2e^y) dy...i am unsure on how to solve that integral on the right side.

 

[LCKurtz]

Hmmm ok, still kind of confused. this is what I did after your correction:

dx/dy + x = 2e^y

I(y) = e^integral (2e^y) dy = e^(2e^y)

multiply integrating factor on both sides will give: e^(2e^y) x = Integral e^(2e^y) (2e^y) dy...i am unsure on how to solve that integral on the right side.

No no. I was just pointing out that your integrating factor is a function of y, not x. As I said before, the solution you had works. You don't need to change it.

 

[HallsofIvy]

So, if I were to solve, would it be like this:

dx/dy + x = 2e^y

I(x) = e^integral dy = e^y

multiply integrating factor on both sides will give: e^y (x) = e^(2y) + C

=> x = e^y + Ce^-y.

That was the answer I got for x, but it does not match answer in the back.

help.

What is the answer in the book? Knowing that might help us!