+267 As the title suggests, you are given two initial conditions, y(0) = 0 and y'(pi)=1
My initial idea was to perform solve this by laplace transform but one of the initial conditions was y'(pi) and not of zero and I don't know how to use that as the laplace transform only uses y'(0).
I assumed that y must be a function of some sort like y = A*sin ax + B*cos bx
After trying to put that into the equation and trying to simplify I ended up with contradicting information regarding the constants and couldn't progress. I tried doing it again and must have made another error somewhere. I would appreciate a walkthrough of this question. Thank you in advance.
+129899 y" + 25y = 3sin2x
First....let's find the solution to the homogenous equation
y" + 25y = 0
The characteristic equation for this is
r^2 + 25 = 0 → r = ± 5i
So, the complementary solution to this homogenous equation is
y = c1 cos(5x) + c2 sin (5x) (1)
Now let's solve the non-homogenous equation
y" + 25y = 3sin2x
Guess that the solution is
y = Acos 2x + Bsin2x
y' = -2Asin2x + 2Bcos2x
y" = -4Acos2x -4Bsin2x
So we have
-4Acos2x - 4Bsin2x + 25(Acos2x + Bsin2x) = 3sin2x
(25A -4A)cos2x + (25B - 4B)sin2x = 3sin2x
21Acos2x + 21Bsin2x = 3sin2x
This implies that A = 0 and B = 1/7
So......the solution to this is that
y = (1/7)sin2x (2)
Combining (1) and (2) we have this
y = c1 cos(5x) + c2 sin (5x) + (1/7)sin(2x)
y' = - 5c1 sin (5x) + 5c2 cos(5x) + (2/7)cos(2x)
Applying the initial conditions
y(0) = c1 (1) = 0 → c1 = 0
y' (pi) = 1
5c2 (-1) + (2/7)(1) = 1
-5c2 + 2/7 = 1
-5c2 = 5/7
c2 = -1/7
So.... the solution with the applied intial conditions is
y = (-1/7)sin(5x) + (1/7) sin(2x)
And
y' = (-5/7)cos(5x) + (2/7)cos(2x)
Check
y(0) = 0 ??
(-1/7)(0) + (1/7)(0) = 0
y'(pi) = 1 ??
(-5/7)(-1) + (2/7)(1)
5/7 + 2/7 = 1
