Question

Use laplace transform to solve : f'(x) - f(x) =e^-x

Answer 1

To solve the differential equation \( f'(x) - f(x) = e^{-x} \) using the Laplace transform, we'll follow these steps:

1. **Take the Laplace Transform of both sides of the equation.**

Let \( F(s) \) be the Laplace transform of \( f(x) \), i.e., \( F(s) = \mathcal{L}\{f(x)\} \).

Using the properties of the Laplace transform, we know:

\[
\mathcal{L}\{f'(x)\} = sF(s) - f(0)
\]

So, taking the Laplace transform of both sides of the equation gives us:

\[
sF(s) - f(0) - F(s) = \mathcal{L}\{e^{-x}\}
\]

The Laplace transform of \( e^{-x} \) is:

\[
\mathcal{L}\{e^{-x}\} = \frac{1}{s + 1}
\]

Therefore, the equation becomes:

\[
sF(s) - f(0) - F(s) = \frac{1}{s + 1}
\]

2. **Rearranging the equation.**

We can have:

\[
(s - 1)F(s) - f(0) = \frac{1}{s + 1}
\]

Solving for \( F(s) \):

\[
(s - 1)F(s) = \frac{1}{s + 1} + f(0)
\]

Thus,

\[
F(s) = \frac{1}{s + 1} \cdot \frac{1}{s - 1} + \frac{f(0)}{s - 1}
\]

3. **Finding the inverse Laplace transform.**

We will separate the terms of \( F(s) \):

\[
F(s) = \frac{1}{(s + 1)(s - 1)} + \frac{f(0)}{s - 1}
\]

To find the inverse Laplace transform, we can use partial fraction decomposition on the first term:

\[
\frac{1}{(s + 1)(s - 1)} = \frac{A}{s - 1} + \frac{B}{s + 1}
\]

Multiplying through by \((s + 1)(s - 1)\) gives:

\[
1 = A(s + 1) + B(s - 1)
\]

Expanding and equating coefficients:

\[
1 = (A + B)s + (A - B)
\]

This gives two equations:

\[
A + B = 0
\]
\[
A - B = 1
\]

Solving these, we get \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \). Thus:

\[
\frac{1}{(s + 1)(s - 1)} = \frac{1/2}{s - 1} - \frac{1/2}{s + 1}
\]

Now substituting back into \( F(s) \):

\[
F(s) = \frac{1/2}{s - 1} - \frac{1/2}{s + 1} + \frac{f(0)}{s - 1}
\]

Combine the terms:

\[
F(s) = \left( \frac{1/2 + f(0)}{s - 1} - \frac{1/2}{s + 1} \right)
\]

4. **Taking the inverse Laplace transform.**

The inverse Laplace transform will give us:

\[
f(x) = \left( \frac{1}{2} + f(0) \right)e^x - \frac{1}{2} e^{-x}
\]

5. **Final Result.**

Thus, the solution to the differential equation depends on the initial condition \( f(0) \):

\[
f(x) = \left( \frac{1}{2} + f(0) \right)e^x - \frac{1}{2} e^{-x}
\]

This gives the general solution for the differential equation \( f'(x) - f(x) = e^{-x} \) in terms of the initial condition \( f(0) \).