Use Laplace Transforms to solve the following differential equation : f’(x)+9f(x) = 2e^2x f(0)=0?
Answers:
Here are a few Laplace Transforms you need to know:
L[f(x)] = F(s)
L[f'(x)] = s*F(s) - f(0)
L[2e^2x] = 2/(s-2)
Now substitute these values into your differential equation:
s*F(s) - f(0) + 9(F(s)) = 2/(s-2). Plug in f(0) = 0 and solve for F(s):
F(s)[s+9] = 2/(s-2) ------------> F(s) = 2/[(s-2)(s+9)]
Now use the inverse Laplace transform to solve for f(x). To do this, you need to transform F(s) into partial fractions. Partial fractions state that:
2/[(s-2)(s+9)] = A/(s-2) + B/(s+9) where A and B are constants. Multiply both sides by the denominator (s-2)(s+9)
2 = A(s+9) + B(s-2) ------->2 = As + 9A + Bs - 2B
Now combine like terms to create equations to solve for A&B
2 = s(A+B) + (9A-2B)
s terms -------------------> A+B=0 (on the left hand side, there are no 's' terms, that's why the equation equals to zero)
constant terms ---------> 9A-2B = 2
Using these two equations, A = 2/11, B = -2/11
Thus,
2/[(s-2)(s+9)] = 2/[11(s-2)] - 2/[11(s+9)]
Now use the inverse Laplace transform:
Since the L^-1[1/(s-a)] = e^(at), we can solve for f(x):
f(x) = (2/11)(e^[2t]) - (2/11)(e^[-9t])
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Below is a website to help you with Laplace transforms. Hope this helps