Verified answer

There is a similar problem worked through by a prof. here

http://www.math.psu.edu/srikrish/math251/notes/dir...

Top quality explanations/examples of inverses: "Paul's online notes"

http://tutorial.math.lamar.edu/Classes/DE/InverseT...

Basic list of transforms here

http://mathworld.wolfram.com/LaplaceTransform.html

It is interesting to note that Dirac delta at t = c is actually the derivative

of the step function at t = c, (makes sense since the graph of

step function is horizontal at every t not equal to c).

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L[f''(t)] = s^2 L[f(t)] - sf(0) - f'(0)

but given that y(0)=y'(0)=0 for this example

L[f''(t)] = s^2 L[f(t)]

L[10] = 10/s

The more familiar L[delta(t)] = 1 is a special case of

L[delta(t - c)] = U[e^-cs] = e^-cs for s > c

where unit step function is given by

U(t) = {0 for t < 0, 1 for t > 0}

and for this example we use

L[10*delta(t - 1)] = 10e^-s (for s > c)

Taking the Laplace transform of your equation

y" = 10+10*delta(t - 1)

s^2 L[f(t)] = 10/s + 10e^-s..............(a)

Pause to note these four Laplace transforms

b) L[U(t)] = 1/s (for real s > 0)

c) L[t*U(t)] = 1/s^2 (for real s > 0)

d) L[U(t - 1)] = e^(-s)/s (for real s > 0)

e) L[t*U(t - 1)] = e^(-s)[1/s^2 + 1/s] (for real s > 0)

Using e - d we obtain

L[(t - 1)*U(t - 1)] =e^(-s)/s^2

L[f(t)] = 10/s^3 + (10/s^2)* e^-s......(f)

Inverse Laplace transform [10/s^3] = 5t^2

Inverse Laplace transform [10e^(-s)/s^2)] = 10[(t - 1)*U(t - 1)]

f(t) = 5t^2 + 10[(t - 1)*U(t - 1)]

This starts of as the parabola but converts to a ramp at t = 1

Regards - Ian