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To solve the differential equation:

\[
y'' + 3y' - 10y = 100x
\]

we proceed by solving the homogeneous equation and then finding a particular solution.

### Step 1: Solve the Homogeneous Equation
The homogeneous equation is:
\[
y'' + 3y' - 10y = 0
\]

This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is:
\[
r^2 + 3r - 10 = 0
\]

Solving for \( r \), we use the quadratic formula:
\[
r = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 10}}{2} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2}
\]

Thus,
\[
r = 2 \quad \text{and} \quad r = -5
\]

So the general solution to the homogeneous equation is:
\[
y_h = C_1 e^{2x} + C_2 e^{-5x}
\]

### Step 2: Find a Particular Solution
For the non-homogeneous part, \(100x\), we assume a particular solution of the form:
\[
y_p = Ax + B
\]

Substitute \(y_p\) into the original equation to determine \(A\) and \(B\):
1. \( y_p' = A \)
2. \( y_p'' = 0 \)

Substituting into the equation \(y'' + 3y' - 10y = 100x\):
\[
0 + 3A - 10(Ax + B) = 100x
\]

Expanding and equating terms:
\[
-10Ax + (3A - 10B) = 100x
\]

To match the terms, we set:
1. \( -10A = 100 \Rightarrow A = -10 \)
2. \( 3(-10) - 10B = 0 \Right

Solve the following differential equation: y'' + 3y' - 10y = 100x

Solve the second-order linear differential equation y'' + 3y' - 10y = 100x with a detailed step-by-step approach. This solution covers finding the homogeneous solution, solving the characteristic equation, and determining a particular solution for non-homogeneous terms. Suitable for college-level students.

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