Initial value problem

C

colerelm

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Oct 24, 2011
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How do I start solving an initial value problem such as this one:

y' = 2y^2 + xy^2, y(0) = 1

I've attempted to solve it but I get stuck at the first step...I try getting it in the form of dy/dx + Py = Q but I don't know how I would get a Q value in this case. This is what I get when I try getting it into the aforementioned form:

dy/dx + y^2*(2+x) = 0

Does that make any sense? Can anyone please step me through the process? Thanks a lot.
 
The ODE associated with the IVP is separable. ;)
 
No, we are given to solve:

\(\displaystyle \dfrac{dy}{dx}=2y^2+xy^2\) where \(\displaystyle y(0)=1\)

Now, if we factor the right side of the ODE, we obtain:

\(\displaystyle \dfrac{dy}{dx}=y^2(x+2)\)

Now divide through by \(\displaystyle y^2\), thereby eliminating the trivial solution \(\displaystyle y\equiv0\), to obtain:

\(\displaystyle \dfrac{1}{y^2}\dfrac{dy}{dx}=x+2\)

Now, integrate with respect to \(\displaystyle x\):

\(\displaystyle \displaystyle \int\frac{1}{y^2}\,dy=\int x+2\,dx\)

Can you proceed from here?
 
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Yeah, I integrate and get -1/y = 2x + (x^2/2) + C. I don't know where to go from there though...
 
colerelm said:
Yeah, I integrate and get -1/y = 2x + (x^2/2) + C. I don't know where to go from there though...
Click to expand...
The problem stated that y(0) = 1,
so if you substitute y=1 and x=0 in the solution, you will have an equation to solve for the explicit value of C.
[Don't neglect the minus sign on the -1/y term.]
 
Ah, that makes sense, thanks. I got C = -1, is that right?

Edit: The solution shows that the answer is:

y(x) = -2/(x^2 + 4x - 2) and x=-2 is the minimum, but how?

I ended up getting y(x) = -2/(2C + x(x+4)) and the minimum being -1...
 
Last edited:
colerelm said:
Ah, that makes sense, thanks. I got C = -1, is that right?

Edit: The solution shows that the answer is:

y(x) = -2/(x^2 + 4x - 2) and x=-2 is the minimum, but how?

I ended up getting y(x) = -2/(2C + x(x+4)) and the minimum being -1...
Click to expand...

Yes, C = -1 is correct. And so we have:

\(\displaystyle \displaystyle -\frac{1}{y}=2x+\frac{x^2}{2}-1=\frac{x^2+4x-2}{2}\)

Hence:

\(\displaystyle \displaystyle y(x)=-\frac{2}{x^2+4x-2}\)

Now, to find any extrema, we equate the derivative to zero:

\(\displaystyle \displaystyle \frac{dy}{dx}=\frac{2(x+2)}{(x^2+4x-2)^2}=0\)

This implies \(\displaystyle x=-2\)

The first derivative test shows that this is a local minimum.
 
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