Question

Solve the given initial value problem and determine at least approximately where the solution is valid.


\((9x^2+y-1)-(4y-x)y\prime=0,~~~y(1)=0\)

Answer 1

I already know that this differential equation is not exact, so, how to solve from there.

Answer 2

Did the question say it's not exact? I can rewrite it by factoring out a negative in the (4y-x) portion.

\[(9x ^{2}+y-1)dx+(x-4y)dy = 0\]

\[\frac{ \delta M }{ \delta y }(9x ^{2}+y-1) = 1\]
\[\frac{ \delta N }{ \delta x }(x-4y) = 1\]

Answer 3

I know where I went wrong, I had:

\(\dfrac{\partial N}{dx}(4y-x)\)

I forgot to distribute the negative to get,

\(\dfrac{\partial N}{dx}(x-4y)\)

Answer 4

Bingo. Then just solve like normal, it works out perfect.

Answer 5

Very cool, thanks for letting me know my silly error :P

Answer 6

No problem :3

Answer 7

Would I use the initial value portion of the problem to solve for C or h(y)?

Answer 8

You dont need the initial value portion to get h(y).

Answer 9

Oh, derp. I feel really really silly right now.

Answer 10

Mhm. Youre just doing a bunch of flip-flopping with which variables you're giving respect to. You do M with respect to y to check for exactness then integrate with respect to x to get h(y), then differentiatr with respect to y again and set that result = to Ny.

So we would have:

\[\int\limits_{}^{}(9x ^{2}+y-1)dx = 3x ^{3}+xy - x + h(y) \]Differentiate with respect to y:

\[\frac{ \delta M }{ \delta y }(3x ^{3}+xy-x+h(y)) = x + h'(y) \]Now this value gets set to N, so we have:

\[x+h'(y) = -4y + x \implies h'(y) = -4y\] Then integrate both sides to get h(y)

\[\int\limits_{}^{}h'(y) = \int\limits_{}^{}-4ydy\]
\[h(y) = -2y ^{2} \] Replace this result in the previous integration and set it all equal to C

\[3x ^{3}+xy - x -2y ^{2}= C\]

Now use the initial value to get C>

If all of that makes sense, lol.

Answer 11

I have:

\(C=3x^3-2y^2+x(y-1)=3x^2-2y^2+xy-1\)

And now that I have this typed, I see you have the same answer.

Answer 12

Which is awesome, good job xD

Answer 13

Thank you, time for another silly question.
\(y(1)=0\)
What form is this?
\(y(x)=?\)

Answer 14

Yeah, pretty much. When x is 1, y is 0 would be the interpretation.

Answer 15

\(2=3x^3+xy-x-2y^2\)
Is this correct?

Answer 16

Thats what I got.

Answer 17

Wait, I need to solve for y....... oh my god.

Answer 18

I suppose you could, yeah, lol. Depends if you have to or not.

Answer 19

According to the back of my book I do..... because I also need to determine at least approximately where the solution is valid.

Answer 20

You know how to go about that then?

Answer 21

Now that I think about it no, my professor never discussed it and looking at the examples, none of them cover it.

Answer 22

Oh, and,

\(y=\dfrac{x-\sqrt{24x^3+x^2-8x-16}}{4}\)

Answer 23

Lol, I didnt even have to write it out it seems xD Or did you look up the answer?

Answer 24

I typed it into my calculator and verified with my book.
I wasn't even going to try with that one.

Answer 25

Well, the way you do it is you treat the x's like constants and then you can say that this is a quadratic in y, which allows you to use the quadratic formula:

So just set everythign equal to 0 and treat it like y^2 + y + c using x's and numbers as constants. I have to go, though, sorry D: Try it out and see if you can get it. Oh, and at the end, the reason its x - instead of +/- is because you need to plug in the intial condition and see which sign makes it true, + or -. But yeah, good luck and glad I could help ^_^

Answer 26

How can you determine approximately where the solution is valid?

Answer 27

Oh well, thank you very very much!