# Math Help - Differential Equations

1. ## Differential Equations

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

y ' = 4y + 4xy.

Can anyone help me with this problem?

2. $4y+4xy=4y(1+x),$ which means that your equation is separable.

Can you take it from there?

3. ## Re: 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:
$y'=4y+4xy=4y(x+1)$
$\frac{1}{y}\,dy=4(x+1)\,dx$
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.

4. ## does this help?

Originally Posted by Jessica098
Given y(2) = 1, estimate y(2.5), where y(x) is the solution of the initial-value problem

y ' = 4y + 4xy.

Can anyone help me with this problem?
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 !