Question

solve ordinary differential equation y''+y=e^x

Answer 1

To solve the ordinary differential equation \( y'' + y = e^x \), we will use the method of undetermined coefficients. This ODE consists of a homogeneous part and a particular part.

1. **Homogeneous Solution:**
First, let's find the solution to the associated homogeneous equation:

\[
y'' + y = 0
\]

The characteristic equation is:

\[
r^2 + 1 = 0
\]

Solving this gives:

\[
r = \pm i
\]

Thus, the general solution to the homogeneous equation is:

\[
y_h = C_1 \cos(x) + C_2 \sin(x)
\]

where \( C_1 \) and \( C_2 \) are constants.

2. **Particular Solution:**
Now, we need to find a particular solution \( y_p \) for the non-homogeneous equation \( y'' + y = e^x \). We will use the method of undetermined coefficients, and we will try a solution in the form:

\[
y_p = A e^x
\]

where \( A \) is a constant to be determined.

We calculate the first and second derivatives of \( y_p \):

\[
y_p' = A e^x
\]
\[
y_p'' = A e^x
\]

Substitute \( y_p \) and its derivatives back into the original equation:

\[
y_p'' + y_p = A e^x + A e^x = 2A e^x
\]

We want this to equal \( e^x \):

\[
2A e^x = e^x
\]

Dividing both sides by \( e^x \) (assuming \( e^x \neq 0 \)) gives:

\[
2A = 1 \implies A = \frac{1}{2}
\]

Therefore, the particular solution is:

\[
y_p = \frac{1}{2} e^x
\]

3. **General Solution:**
The general solution \( y \) to the original non-homogeneous equation is the sum of the homogeneous and particular solutions:

\[
y = y_h + y_p = C_1 \cos(x) + C_2 \sin(x) + \frac{1}{2} e^x
\]

Thus, the final solution to the ordinary differential equation \( y'' + y = e^x \) is:

\[
\boxed{y = C_1 \cos(x) + C_2 \sin(x) + \frac{1}{2} e^x}
\]

where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial or boundary conditions.