[ODEs] Exact Equations and Substitution

[FermiParadox]

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.

Problem 1: Exact Equation

Homework Statement

Given the differential equation (12+5xy)dx + (6xy^-1+3x²)dy=0 which is not exact, find an integrating factor (in the form xⁿy^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 a wrong general solution.

Homework Equations

(12+5xy)dx +(6xy^-1+3x²)dy = 0
u = xⁿy^m

The Attempt at a Solution

a. Multiply through by xⁿy^m, giving:

(12xⁿy^m+5xⁿ⁺¹y^m+1) dx + (6xⁿ⁺¹y^m-1+3xⁿ⁺²y^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​

[12bxⁿy^m-1 + 5xⁿ⁺¹y^m(m+1)] = [6xⁿy^m-1(n+1)+3xⁿ⁺¹y^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 xⁿy^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 x³y², yet apparently that's wrong. What gives?

Problem 2: Substitution

Homework Statement

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).

Homework Equations

y' = f(ax+by+c)
f(x,y) = C
y' = 1-e^y-x+1

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^-vdv = 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.

 

[dynamicsolo]

Problem 1: Exact Equation

So our integrating factor is x³y², yet apparently that's wrong. What gives?

I don't see the problem either on this one (I though I'd found a sign error for m at first, but it turned out to be mine...). Multiplying through by x³y² does make M_y = N_x .

What exactly does the input request? The integrating factor itself or just the exponents of x and y? (Have I said how much I hate computer-based problem systems? Well, maybe only in every twentieth thread...)

 

[FermiParadox]

Yeah, I hate them too. It's convenient when I want to miss class and still do the homework, I guess, and at least this one doesn't penalize for multiple tries, but I usually spend more time figuring out the right syntax than I do solving the problem.

The input calls for the integrating factor - I think there might just be a bug on this particular problem because I went ahead and solved the stupid thing and got a 50% on it after inputting the solution, so... not really sure.

 

[dynamicsolo]

It could be that the stored answer that the computer compares student answers to has a typo; that's been known to happen. I wonder if the instructor forgot the order of the exponents and entered "x²y³", but nonetheless solved the DE correctly.

I can look at #2 in a while...

 

[FermiParadox]

Thanks, I'd appreciate that - and thank you again for the help so far.

 

[dynamicsolo]

Problem 2: Substitution
...

v' = -e^v
-e^-vdv = dx[/indent]

c. Integrate both sides, substitute back in v.

e^-v = x+C
e^-v-x = C
e^-(y-x+1)-x=C​

The only problem I can see there might be is that you don't have quite the form the computer was given as "correct". But this does solve the original DE -- consider by implicit differentiation:

\frac{d}{dx} [e^{-(y-x+1)} - x ] = \frac{d}{dx} C \Rightarrow [-e^{-(y-x+1)} \cdot ( y' - 1 ) ] - 1 = 0 ,

using a bit o' Chain Rule, so

\Rightarrow [-e^{-(y-x+1)} \cdot ( y' - 1 ) ] = 1 \Rightarrow y' - 1 = \frac{1}{-e^{-(y-x+1)}} = -e^{(y-x+1)} ,

\Rightarrow y' = 1 - e^{(y-x+1)} ,

which is the DE we started with.

By doing the problems on the computer system thoroughly, you may just be finding all the "buried, unexploded ordinance" in the answer database. How does it feel to be a minesweeper...?