Differential Equations. Reduction of Order. Calc IV.

[Gummy Bear]

Homework Statement

(x²+2y')y''+2xy'=0 with initial conditions of y(0)=1 and y'(0)=0

Homework Equations

reduction of order, and exact equations..

The Attempt at a Solution

let y'=p and y''=p'
Giving: (x²+2p)p'+2xp Let M=2xp and N=(x²+2p)
So, M_p=2x and N_x=2x Therefore, they are exact.
Let f_x=2xp
so, f(x,p)=x²p+h(p)
f_p=x²=h'(p)
comparing to f_p=x²+2p we know h'(p)=2p so, h(p)=p²
so, c=p²+px²
Now, solve for p: using quadratic.
p=(-x²±√(x⁴+4c))/2
Substitute back for y'=p
y'=(1/2)(-x²±√(x⁴+4c)
y=?

This is where I'm not sure what to do.

I'm pretty sure I use the initial conditions, do I assume c=0 and take the integral of:
(1/2)(-x²±√(x⁴))

 

[LCKurtz]

Homework Statement

(x²+2y')y''+2xy'=0 with initial conditions of y(0)=1 and y'(0)=0

Homework Equations

reduction of order, and exact equations..

The Attempt at a Solution

let y'=p and y''=p'
Giving: (x²+2p)p'+2xp Let M=2xp and N=(x²+2p)
So, M_p=2x and N_x=2x Therefore, they are exact.
Let f_x=2xp
so, f(x,p)=x²p+h(p)
f_p=x²=h'(p)
comparing to f_p=x²+2p we know h'(p)=2p so, h(p)=p²
so, c=p²+px²

Right there you have $y'^2+x^2y'=c$. If you use $y'(0)=0$ you get $c=0$ so your equation is $y'^2+x^2y'=y'(y'+x^2)=0$. This tells you that either $y'=0$ or $y'=-x^2$, both of which are easy to handle.

 

[Gummy Bear]

I'm confused, you're substituting back for p before solving for p? Also, y'(0)≠0 y'(0)=1.

Does that mean my equation should be:
y'²+x²y'=1

And from there solve for y?

 

[LCKurtz]

Homework Statement

(x²+2y')y''+2xy'=0 with initial conditions of y(0)=1 and y'(0)=0

Right there you have $y'^2+x^2y'=c$. If you use $y'(0)=0$ you get $c=0$ so your equation is $y'^2+x^2y'=y'(y'+x^2)=0$. This tells you that either $y'=0$ or $y'=-x^2$, both of which are easy to handle.

I'm confused, you're substituting back for p before solving for p? Also, y'(0)≠0 y'(0)=1.

I have quoted your original boundary condition. $p=y'$ so I just put it in. It doesn't matter which you call it, but using $y'$ emphasizes that is what it is, and it makes it more natural when using the boundary conditions.

Does that mean my equation should be:
y'²+x²y'=1

No. You get 0 on the right side. It means either $y'=0$ or $y'=-x^2$. Solve those, apply any boundary conditions and, if you aren't sure, check the solutions work.

 

[Gummy Bear]

Wow. I got it mixed up with the problem out of the book. It's been a long day. You're right..

So all I have to do is:

y=∫0dx=c

y=∫(-x²)dx=(-x³)/3+c

Therefore, y=∫(-x²)dx=(-x³)/3+c is the answer?

Thanks for helping me by the way.

 

[LCKurtz]

Wow. I got it mixed up with the problem out of the book. It's been a long day. You're right..

So all I have to do is:

y=∫0dx=c

y=∫(-x²)dx=(-x³)/3+c

Therefore, y=∫(-x²)dx=(-x³)/3+c is the answer?

Thanks for helping me by the way.

Use the boundary condition to evaluate c. A second order equation with two initial conditions won't have any arbitrary constants left. And you shouldn't use the same c for both constants.

 

[LCKurtz]

I should add here that this is a non-linear equation. You get two separate solutions. Nothing says that the sum of those two solutions is also a solution.

 

[Gummy Bear]

Oh, so now I say because y(0)=1 0/3+c₂=1 so, c₂=1?

Therefore, y=-x³/3+1

What do you mean don't use the same constant? Are you referring to y=∫0dx=c₁?

What do I do with the y=∫0dx=c₁ anyway?

 

[LCKurtz]

Oh, so now I say because y(0)=1 0/3+c₂=1 so, c₂=1?

Therefore, y=-x³/3+1

What do you mean don't use the same constant? Are you referring to y=∫0dx=c₁?

What do I do with the y=∫0dx=c₁ anyway?

You apply the boundary condition $y(0)=1$. Remember you used $y'(0)=0$ to get the two separate solutions. Each of them satisfies $y(0)=1$. So you have two different solutions to this problem. You should check that they both work.

 

[Gummy Bear]

Ya, I figured that out after asking. But how do I check that both -x³/3+1 and 1 work?

 

[LCKurtz]

Ya, I figured that out after asking. But how do I check that both -x³/3+1 and 1 work?

Plug them into the equation and boundary conditions.