How do you solve the differentail equation y' = x + y?

If you can tell me the application you’re using it for, I’ll help you.

Its just a homework problem that the professor has not explained how to solve yet, no actual application.

Yeah, the application might help.

the question is “Solve the differential equation y’ = x + y by making the variable substitution u = x + y” that’s all I have to go on.

I know someone is going to come in here and make me feel stupit, but what exactly is y’ and y”?

Try this… does it help?

y’ is the derrivative of the y, which is a function of x, and that link does not help.

Taking your algebra one step further, you have u = du – 1, or u – du + 1 = 0.

Try u = e^x – 1

Ok, it’s not the exact same equation, but maybe this one from Yahoo! Answers will help:
y’=y/x

dy/dx = y/x

dy/y = dx/x

Integrating,

Iny = Inx + c

Iny = Inx + InC

Iny = In{xC}

y = Cx, where C is a constant.

A differential equation of the form y’ + p(t) y = x(t) can be solved by finding the integrating factor μ(x). Look in your notes or your book for a complete proof of this, but this factor can be found with the following equation.

μ(t) = e ^ (∫ p(t) dt)

The purpose of μ(t) is to satisfy μ’(t) = μ(t) p(μ). Multiplying the integrating factor by each term will give you μ(t) y’ + μ(t) p(t) y = μ(t) x(t) which is μ(t) y’ + μ’(t) y = μ(t) x(t) by the previously stated property of μ. The LHS is simply the product property of derivatives, so it can be written as (μ(t) y(t))’ = x(t). This allows you to integrate each side, so you can easily solve for y(t) using algebra.

The integrating factor of your equation is μ(t) = e ^ (∫ -1 dt) = e^-t, so you can write the problem as (e^-t y(t))’ = x(t). Thus, the final answer is y(t) = e^t ∫ x(t) dt. You’re welcome.

Please note that I’m completely ignoring integration constants. Your professor will want these, so it would do you good to rework this with the constants now that I’ve led you through the process.

Note: It has been a long time since I’ve done this and I just realized how rusty I am. I think this is correct, but it looks different from what Vortico got so now I’m not sure. I can’t believe I forgot this!

This is called the “u-substitution” method. Look in your text for more info.

The basic point of it is to substitute a third variable, u, for a key function to make integration much easier. In your case,

dy/dx = x+y ; u = x + y

Therefore:

dy/dx = u and by rearranging,

dy = u dx

Then find the replacement for dx from the relationship u = x + y:

du = dx ( y is treated as a constant when diff w/ respect to x )

Now make the substitution for dx:

dy = u du

Now integration is simple and straight forward:

y = u^2/2 + C

Once you integrate to solve for y, substitute “x+y” back into the equation for u:

y = (x+y)^2/2 + C

Okay by differentiating my solution, I can verify it is correct -

1. d/dx to both sides: d/dx (y) = d/dx [ (x+y)^2/2 + C ]

2. Pull out the factor of 0.5: dy/dx = (½) d/dx [ (x+y)^2 + C ]

3. Recognize that d/dx C = 0: dy/dx = (½) d/dx [ (x+y)^2 ]

4. Multiply out (x+y)^2: dy/dx = (½) d/dx [ x^2 + 2xy + y^2 ]

5. Take the derivative of each term: dy/dx = (½) [ 2x + 2y ]

Note: d/dx x^2 = 2x; d/dx 2xy = 2y since y is treated as a constant and deriv of x is 1; d/dx y^2 = 0 since y is treated as constant.

6. Simplify: dy/dx = x + y

How do I solve it? Put it into wolfram alpha:
http://www.wolframalpha.com/input/?i=y'+=+x+-+-y

(many edits to figure out something that would work with fluther’s screwy formatting)

@phaedryx Many times the answer is in the back of the book. Wolfram is great for answers, but the solution is what the OP is looking for.

Yeah, it sounded like s/he didn’t have a lot of information about the problem. I figured the solution might help.

Okay I think I missed the fact that this is a linear ODE…Here’s a solution I think is right:

dy/dx = x + y

GENERAL FORM:

dy/dx + H(x) y + Z(x) = 0; H(x) = -1 and Z(x) = -x

THEREFORE:

dy/dx – y – x = 0

MULTIPLY THROUGH INTEGRATING FACTOR e^∫ [ H(x) dx ]=e^ (-∫dx)=e^(-x+C)=Ce^(-x):

(dy/dx)Ce^(-x) – yCe^(-x) – xCe^(-x) = 0

YOU CAN PROVE d/dx [ yCe^(-x) ] = (dy/dx)Ce^(-x) – yCe^(-x) VIA CHAIN RULE:

d/dx ( yCe^(-x) ) = -yCe^(-x) + Ce(-x) dy/dx

THEREFORE SUBSTITUTING THE FORMER FOR THE LATTER:

d/dx ( yCe^(-x) ) – xCe^(-x) = 0

INTEGRATE BOTH SIDES:

∫ [d/dx ( yCe^(-x) ) dx – ∫ [xCe^(-x)] dx = 0

SIMPLIFY:

yCe^(-x) – C∫ [xe^(-x)] dx = 0

REARRANGE:

yCe^(-x) = C∫ [xe^(-x)] dx

SIMPLIFY:

ye^(-x) = ∫ [xe^(-x)] dx

y = e^x ∫ [xe^(-x)] dx

y = e^x [ – e^(-x) (x+1) + D ]; where D is a constant

y = e^x [ -e^(-x)x -e^(-x) + D ]

y = – x e^x e^(-x) – e^x e^(-x) + D e^x

y = -x -1 + De^x

y = De^x – x – 1

Can anyone verify as it has been a long time since I’ve done anything like this…And what’s with the reference in the question to u = x + y?

Correct for upstairs, the chain rule shows that – yCe^(-x) is zero. d/dx ( yCe^(-x) ) – xCe^(-x) = 0 follows, and ∫ [xe^(-x)] dx is by parts. this method was mentioned above as solving it by integrating factor, method to solve homogeneous, ODEs like y’=x+y.