SOLUTION: please help me answer this question: Determine the particular solution of the equation, {{{ d^2y/dx^2 + 4(dy/dx)= e^(-2x) }}} given the initial conditions, y(0)=0, y'(0)=0

Algebra ->  Customizable Word Problem Solvers  -> Evaluation -> SOLUTION: please help me answer this question: Determine the particular solution of the equation, {{{ d^2y/dx^2 + 4(dy/dx)= e^(-2x) }}} given the initial conditions, y(0)=0, y'(0)=0       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1115566: please help me answer this question:
Determine the particular solution of the equation,
+d%5E2y%2Fdx%5E2+%2B+4%28dy%2Fdx%29=+e%5E%28-2x%29+
given the initial conditions,
y(0)=0, y'(0)=0

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
I will solve for the general solution, using Laplace transforms.

L{y"} = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) (because y'(0)=y(0)=0)
L{y'} = sY(s) - y(0) = sY(s)
L{e^(-2x)} = 1/(s+2)

Now in the s-domain, we can write:
s%5E2Y%28s%29+%2B+4sY%28s%29+=+1%2F%28s%2B2%29+
Solving for Y(s):
+Y%28s%29+=+1%2F%28%28s%2B2%29%28s%5E2%2B4s%29%29+=+1%2F%28%28s%29%28s%2B2%29%28s%2B4%29%29+

Using partial fraction expansion:
+1%2F%28%28s%29%28s%2B2%29%28s%2B4%29%29++=+A%2Fs+%2B+B%2F%28s%2B2%29+%2B+C%2F%28s%2B4%29+

Multiplying both sides by (s)(s+2)(s+4) gives us:
+A%28s%2B2%29%28s%2B4%29+%2B+B%28s%29%28s%2B4%29+%2B+C%28s%29%28s%2B2%29+=+1+
+A%28s%5E2%2B6s%2B8%29+%2B+B%28s%5E2%2B4s%29+%2B+C%28s%5E2%2B2s%29+=+1+

This generates 3 equations in 3 unknowns:
from s%5E2: A+B+C = 0 (1)
from s%5E1: 6A+4B+2C = 0 (2)
from s%5E0: 8A = 1 (3) —> A=1/8
Multiply (1) by 2 and subtract (2) to get: -4A-2B = 0, plug in A=1/8 to get B= -1/4
Now plugging in A=1/8, B=-1/4 into (1) and solving for C: C=1/8
Thus:


Finally, +y%28x%29+=+L%5E-1%28Y%28s%29%29+:



Check:
y(0) = 1/8 - 1/4 + 1/8 = 0 (ok)
y'(x) = +%281%2F2%29e%5E%28-2x%29+-%281%2F2%29e%5E%28-4x%29+ and y'(0) = 1/2 - 1/2 = 0 (ok)
y''(x) = +-e%5E%28-2x%29+%2B+2e%5E%28-4x%29+
y'' + 4y' =
= +++-e%5E%28-2x%29+%2B+2e%5E%28-4x%29+%2B+2e%5E%28-2x%29+-+2e%5E%28-4x%29+
= ++e%5E%28-2x%29+ (ok, whew!)