Question

Differential Equations - just wanted to verify my work.

Solve the following IVP ;

(3x^2+y-2)dx + (x+2y)dy = 0
y(2) = 3

My = 1
Nx = 1

f(x,y) = Integral(M(dx) + g(y))

= Integral(3x^2 + y - 2)dx + g(y)
= x^3 - 2x + g'(y)

x^3 - 2x + g'(y) = x + 2y
x^3 + g'(y) = 3x + 2y
g'(y) = Integral(3x + 2y - x^3)dy

g(y) = y^2

f(x,y) = x^3 -2x + g(y)
C = x^3 - 2x + y^2

y(2) = 3

C = (2)^3 - 2(2) + (3)^2
C = 8 - 4 + 9
C = 13

13 = x^3 - 2x + y^2

0 = x^3 -2x +y^2 -13

Just wanted to verify that all of the steps I took were correct! Thanks!

Answer 1

Integral(3x^2 + y - 2)dx + g(y)
= x^3 + xy -2x + g(y)

Answer 2

and totaly don't understand this:
x^3 - 2x + g'(y) = x + 2y
if you take derivative respect to y that's not what you get, :)

Answer 3

you there? @Archetype

Answer 4

Yes. One second.

Answer 5

If you whant i explain you in detail this type of ODE

Answer 6

I believe I left off the integral sign? Everything you typed looked fine to me... I didn't take the derivative with respect to y.

Answer 7

= Integral(3x^2 + y - 2)dx + g(y)
Derivative with respect to X, which gives this;

= x^3 - 2x + g'(y)

Answer 8

then you wrong, my friend ..:)

Answer 9

How so? Where did I go wrong? Guess I do need an explanation.

Answer 10

If I take the integral of X in the equation (x + y) then Y is treated as a constant and is = 0.

Answer 11

Ok. This type of ODE is colled exact differ. eq.. It means that the equation is actualy a full differential of some function F(x,y), whose partial derivatives respect to x, Mx, and y, My are whats stands by dx and dy in the equation respectivly. Till now, ok?

Answer 12

To actualy check that we got this type of equation is the step of checking the second derivatives are equal... That was your 1º step

Answer 13

My=Nx

Answer 14

Yeah.. I know the basics of it. I believe my work should show that. If I am making a mistake it should be something fairly simple.

Answer 15

not really, :)

Answer 16

Integral(3x^2 + y - 2)dx + g(y)
= x^3 + xy -2x + g(y)

From what i have read; that is not correct.

Answer 17

Unless I'm just out to lunch. I imagine you have far more experience with these then I do.

Answer 18

just keep in mind that M is partial derivative of F respect to x. And N is Fy

Answer 19

So to find F you can integrate Fx, which you call M respect to x. So lets do it:
Fx = 3x^2 + y - 2
keep in mind that now we treat y like a constant parameter.
\[F = \int\limits (3x ^{2} +y -2) dx= x ^{3}+xy -2x +g(y)\]
now, if we would take derivative respect to y of this result it should be equal to N. So let's do it:
x+g'(y) = N = x+2y
so g'(y) = 2y
g=y^2 +C
To find C use the information y(2) = 3
Ok?

Answer 20

got it?

Answer 21

Yes.. but it looks like my only mistake was integrating 3x^2 + y -2 incorrectly?

Answer 22

notonly, you didn't take derivative respect to y, you did respect to x....

Answer 23

So this, for example... Integral of (7x-3y)dx = (7/2)x^2 + 3yx ?

Answer 24

yes

Answer 25

Ugh, ok, excellent. Well thank you very much for the help. I've made that mistake on every exact DE problem I've done so far.

Answer 26

glad that could help

Answer 27

Ugh, yeah, just typed it into Wolframalpha and it gave the same (not that i doubted you..) Thanks again mate, you were a GREAT help!

Answer 28

yw