Calculus 2: Solve the initial value problem?

Question

y' + y = 2e^x; y(0) = 8

I think I did this horrible wrong. I used Integration Factoring and got: y(x) = 2x/e^x + 2C/e^x

Anyone know how to do this?

Thanks,
Derek

Answer 1

It´s easier The caracteristic equation is r+1=0 so r=-1
The solution of the equation without second side is
y= C*e^-x
A particular solution with secon side is y=e ^x
so the genera<
la solution is
y=Ce^-x+e^x
At x=0 8=C+1 so C=7 and
y=7e^-x+e^x

Answer 2

i'm assuming that the expression is dy/dx = sec²(x) + 2sin(x) the place y(?/4) = a million. via separation of variables: dy = sec²(x) + 2sin(x) dx Then, ? dy = ? sec²(x) + 2sin(x) dx ? dy = ? sec²(x) + 2 ? sin(x) dx y = tan(x) - 2cos(x) + c If y(?/4) = a million, then... a million = tan(?/4) - 2cos(?/4) + c a million = a million - 2(?(2)/2) + c a million = a million - ?(2) + c a million - a million + ?(2) = c c = ?(2) ? y = tan(x) - 2cos(x) + ?2 i'm hoping this helps!