Giveny(2) = 1, estimatey(2.5), wherey(x) is the solution of the initial-value problem

y' = 4y+ 4xy.

Can anyone help me with this problem?

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- Jun 20th 2008, 03:40 PMJessica098Differential Equations
Given

*y*(2) = 1, estimate*y*(2.5), where*y*(*x*) is the solution of the initial-value problem

*y*' = 4*y*+ 4*x**y*.

Can anyone help me with this problem? - Jun 20th 2008, 04:07 PMKrizalid
$\displaystyle 4y+4xy=4y(1+x),$ which means that your equation is separable.

Can you take it from there? - Jun 20th 2008, 04:07 PMTwistedOne151Re: Estimation
Since it says "estimate y(2.5)," you could use a numerical method like the Euler method to obtain an estimate.

Or, you could use the fact that the differential equation is separable:

$\displaystyle y'=4y+4xy=4y(x+1)$

$\displaystyle \frac{1}{y}\,dy=4(x+1)\,dx$

to solve for y(x) explicitly, and then plug in x=2.5

--Kevin C.

Edit: I see Krizalid beat me to pointing out the separability of the equation. - Oct 27th 2008, 08:29 PMjf314does this help?
If you write the differential equation as

dy/dx - (4 + 4x) y = 0

then find the integrating factor of mu(x) = exp(-4x-2x^2) then mult both sides by mu(x) you get

d/dx [ mu(x) y(x) ] = 0 or mu(x) y(x) = c ... then solve for y(x) you get

y(x) = c exp(4x+2x^2) ... so plugging this in the IC y(2) =1 you get

1 = c exp(16) or c = exp(-16) .. hence the solution is

y(x) = exp(-16) exp(4x + 2x^2) so y(5/2) = exp(-16) exp (65/2) = exp(33/2) (if i did my arithmetic right) ... thats a BIG number ... that DE goes up awfully fast !