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 \( f(0) \).