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PG,
For 0 ≤ x ≤ 3, y'+2y=1, so the integrating factor is e^∫2dx = e^2x. That gives y'e^2x+2e^2x = e^2x, or (ye^2x)' = e^2x. Integrating both sides gives ye^2x = e^2x/2 + c or y = ½ + ce^-2x. Substituting the initial conditions gives 0 = ½+c ==> c = -½. So we have y = ½ -½e^-2x for 0 ≤ x ≤ 3.
If x is outside the above range, y'+2y=0, the variables are separable, and the solution is easy.
Question. Did you use Snagit to write your request. If not, what processor did you use.
Ratch
MrAl,
As I understand the problem, if x > 3, then the differential equation is y' + 2y = 0 ==> y' = -2y ===> dy/y = -2xdx. Integrating both sides gives ln(y) = -x² + c ===> y = e^-2x²+e^c = ce^-2x². The initial condition cannot be applied here because y(0) is not defined for this solution due to x > 3.
Ratch
I think you meant this though:
dy/y=-2*dx
so:
ln(y)=-2*x+c
so:
y=e^(-2*x+c)
Right?
Interestingly, the new initial condition can come from the first equation solved for 0<=x<=3 . Since we solved for y we can set x=3 and get the new initial condition.
If you'd like to try this solution it might be good to see the complete procedure here.