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.

Quazars
Apr 20, 2017

#2**+3 **

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

CPhill
Apr 20, 2017