Okay, I'm going insane. I have these problems completely worked out and have stared at them for centuries but the online homework is still telling me they're wrong. Could anyone here take a look and let me know? I'd appreciate it a ton.(adsbygoogle = window.adsbygoogle || []).push({});

Problem 1: Exact Equation

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

Given the differential equation (12+5xy)dx + (6xy^{-1}+3x^{2})dy=0 which is not exact, find an integrating factor (in the form x^{n}y^{m}) and use it to find the general (nonzero) solution. I'm not worried about the general solution because one of the input fields calls for finding the integrating factor and because it tells me I'm wrong, I haven't bothered to go any further to find awronggeneral solution.

2. Relevant equations

(12+5xy)dx +(6xy^{-1}+3x^{2})dy = 0

u = x^{n}y^{m}

3. The attempt at a solution

a. Multiply through by x^{n}y^{m}, giving:

(12x^{n}y^{m}+5x^{n+1}y^{m+1}) dx + (6x^{n+1}y^{m-1}+3x^{n+2}y^{m})dy = 0

b. Now that the equation has been multiplied by the integrating factor, the test for exactness should come up true. Therefore,

M_{y}= N_{x}[12bx^{n}y^{m-1}+ 5x^{n+1}y^{m}(m+1)] = [6x^{n}y^{m-1}(n+1)+3x^{n+1}y^{m}(n+2)]

c. Okay, now that we have all of those we should be clear to set like terms - those with the same power of x^{n}y^{m}- equal to one another, and then solve for n and m.

12m = 6(n+1)

12m = 6n+6

12m-6 = 6n

2m-1 = n

5(m+1) = 3(n+2)

5m+5 = 3n + 6

5m = 3n+1

5m = 3(2m-1)+1

5m = 6m-2

-m = -2

m = 2

n = 2m-1

n = 2(2)-1

n = 3

So our integrating factor is x^{3}y^{2}, yet apparently that's wrong. What gives?

Problem 2: Substitution

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

Find the general solution of

y' = 1-e^{y-x+1}

Notice the right-hand-side is in the form y' = f(ax+by+c).

The solution should be written in the form f(x,y) = C (where C is the constant of integration).

2. Relevant equations

y' = f(ax+by+c)

f(x,y) = C

y' = 1-e^{y-x+1}

3. The attempt at a solution

a. Fairly straightforward substitution. I decided to do

v = y-x+1

v' = y'-1

y' = v'+1

b. Now plugging that into the equation, we get-

v'+1 = 1-e^{v}

v' = -e^{v}

-e^{-v}dv = dx

c. Integrate both sides, substitute back in v.

e^{-v}= x+C

e^{-v}-x = C

e^{-(y-x+1)}-x=C

Not sure what I did wrong here, either. I have a habit when progressing through these problems of making tiny mistakes that botch the entire thing, so if I did that here, please let me know! I think my general approach is right, though.

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# Homework Help: [ODEs] Exact Equations and Substitution

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