SOLUTION: Please help me answer the following question:
(a) Determine the particular solution of the equation
{{{ d^2y/dx^2 - 3(dy/dx) = 9 }}}
given the initial conditions:
y(0) =
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Question 1115567: Please help me answer the following question:
(a) Determine the particular solution of the equation
given the initial conditions:
y(0) = 0, y'(0) = 0
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!
Using Laplace transforms:
L(y'') = Y(s) - sy(0) - y'(0)
L(y') = sY(s) - y(0)
L(c) = c/s (c=a constant)
Noting y'(0)=y(0)=0, the Laplace transform is:
Now we "just" need to isolate Y(s) and then take the inverse Laplace transform.
Use partial fraction expansion to get the right hand side into a form in which the inverse Laplace can be taken:
(1)
Multiply both sides by :
From this you get three equations in three unknowns:
A+C = 0 (from the terms)
-3A+B = 0 (from the terms)
-3B = 9 (from the terms)
> B = -3 > A= -1 > C = 1
So we can write (1) as:
Taking the inverse Laplace gives:
Check:
Initial conditions:
y(0) = -1-3*0+e^(0) = -1 + 1 = 0 (ok)
y'(x) = -3 + 3e^(3x) and y'(0) = -3 + 3e^(0) = -3+3 = 0 (also ok)
Entire equation:
y''(x) = 9e^(3x)
y'' - 3y' = 9e^(3x) - 3(-3+3e^(3x)) = 9e^(3x) + 9 -9e^(3x) = 9 (ok)
My answer is the general solution. A particular solution is often guessed at the start, and then combined with the homogeneous solution (i.e. particular solution would be a function y(x) that satisfies y''-3y' = 9 while the homogenous ("complementary") solution would satisfy y''-3y' = 0 and you add the two solutions together to get the general solution. I don't know how to guess a proper particular solution for this problem. One could guess y(x) = Ae^(kx) + Bxe^(mx) + C, I suppose, but I wouldn't know to guess that without seeing the general solution first.
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