Question

find the inverse laplace transform of 1/(s+1)(s+2)

Answer 1

Answer:

see image for solution.

Answer 2

The inverse Laplace transformation of f(s) is

$\bf \: f(t) = {e}^{ - t} - {e}^{ - 2t}$

Given:

• $f(s) = \frac{1}{(s + 1)(s + 2)}$

To find:

• Inverse Laplace transformation of f(s).

Solution:

Concept/Formula to be used:

• Do a partial fraction of f(s).
• Partial fraction: $f(x) = \frac{1}{(x + a)(x + b)} = \frac{A}{x + a} + \frac{B}{x + b}$
• Write an inverse transformation of f(s).
• $\text{Inverse Laplace transformation of} \: \left( \frac{1}{x - a} \right) = {e}^{at}$

Step 1:

Do the partial fraction of f(s).

$\frac{1}{(s + 1)(s + 2)} = \frac{A}{s + 1} + \frac{B}{s + 2}$

Find the values of A and B.

$\frac{1}{(s + 1)(s + 2)} = \frac{A(s + 2) + B(s + 1)}{(s + 1)(s + 2)}$

$\frac{1}{(s + 1)(s + 2)} = \frac{As + 2A + Bs + B}{(s + 1)(s + 2)}$

compare the coefficients of 's' and constant term.

$A+B= 0 ...eq1$

$2A + B = 1...eq2$

Subtract eq2 from eq1.

$A+B= 0 \\ 2A+ B = 1 \\ ( - ) \: ( - ) \: ( - ) \\ - - - - - - - \\ - A = - 1$

$\bf A = 1$

Put the value of A in eq 1.

$\bf B = - 1$

So,

The function can be written as

$\bf = \frac{1}{s + 1} - \frac{1}{s + 2}$

Step 2:

Write the inverse Laplace transformation of f(s).

${L}^{ - 1}f(s) ={L}^{ - 1} \left( \frac{1}{s + 1} \right) - {L}^{ - 1} \left( \frac{1}{s + 2} \right)$

$\bf {L}^{ - 1}f(s) = {e}^{ - t} - {e}^{ - 2t}$

Thus,

The inverse Laplace transformation of f(s) is $\bf f(t) = {e}^{ - t} - {e}^{ - 2t}$

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