What is the Solution to a Differential Equation with Initial Conditions?

[pat666]

Homework Statement

$$(2x*y-5)dx+(x^2+y^2)dy=0$$ test for exactness and solve

Homework Equations

The Attempt at a Solution

I've been following an example to try and solve this but I'm not sure if its right or how to finish it of.
$$\delta M/\delta y=2x$$
$$\delta N/\delta x=2x$$
therefore it is exact

$$u=x^2y-5x+k(y)$$
$$\delta y/\delta y =x^2+k'=N=x^2+y^2$$
$$k=y^3/3 +c$$

$$u=x^2y-5x+y^3/3 +c$$

initial conditions were y(3)=1 but I still can't find c because I have u x y and c?

not even sure if its right up to this point?

Thanks for any help!

 

[HallsofIvy]

Where did "u" come from? Your original equation only involved x and y!

Yes, I know what you are doing- the fact that f(x,y)dx+ g(x,y)dy is an "exact" equation means that there exist a function, u(x,y), such that du= f(x,y)dx+ g(x,y)dy.

But my point is that saying that du= f(x,y)dx+ g(x,y)dy= 0 means that u is a constant. That is, you can drop "u" completely and write your solution as
$$x^2y- 5x+ y^3/3= c$$
Now, replace x with 3 and y with 1
to solve for c.

 

[pat666]

so y=x^2-5x+y^3/3-17/3??

Thanks

 

[HallsofIvy]

How in the world did you get that? Did you simply ignore what I said?
The solution is
$$x^2y- 5x+ y^3/3= c$$
for some number c.

Put x= 3, and y= 1 into that to determine c.

 

[pat666]

thats what I did and I got c= -17/3 then I just moved it. I'm just learning odes so I wasn't sure how to leave the final form.
$$x^2y- 5x+ y^3/3= -17/3$$ ?

thanks

 

[HallsofIvy]

No, you did not just "move it". You can write the solution as
$$x^2y- 5x+ y^3/3= -17/3$$
or as
$$x^2y- 5x+ y^3/3+ 17/3= 0$$
or, if you don't like fractions, as
$$3x^2y- 15x+ y^3+ 17= 0$$

But none of those is the same as
$$y=x^2-5x+y^3/3-17/3$$

 

[pat666]

oops, no there not that was a silly mistake.

Thanks.