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# Solve the following differential equation: y''+25y = 3sin2x

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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
edited by Quazars  Apr 20, 2017
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#1
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Did you try to see if Wolfram/Alpha will give you an answer?

Guest Apr 20, 2017
#2
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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
edited by CPhill  Apr 21, 2017
#3
+245
+1

I'm sure I say this every time but you are a life saver. Thank you :)

Quazars  Apr 21, 2017

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