This tutorial explains how to solve second order differential equations when the auxiliary equation has two equal real solutions. Step-by-step examples with detailed solutions are included.
General Method
Consider a second order differential equation with constant coefficients:
\[ \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = 0 \]Its auxiliary equation is:
\[ k^2 + b k + c = 0 \]If the discriminant satisfies:
\[ b^2 - 4c = 0 \]then the equation has two equal real solutions, given by:
\[ k = -\frac{b}{2} \]In this case, the general solution of the differential equation is:
where \( A \) and \( B \) are constants determined by initial conditions.
Verification
Let \( y = x e^{kx} \). Then the derivatives are:
\[ y' = e^{kx} + k x e^{kx}, \quad y'' = 2 k e^{kx} + k^2 x e^{kx} \]Substituting into the differential equation:
\[ y'' + b y' + c y = 2 k e^{kx} + k^2 x e^{kx} + b(e^{kx} + k x e^{kx}) + c(x e^{kx}) \]Factor and simplify:
\[ e^{kx} \left[ 2k + b + x (k^2 + b k + c) \right] = 0 \]Since \( k^2 + b k + c = 0 \) and \( 2k + b = 0 \) when \( k = -\frac{b}{2} \), the solution satisfies the differential equation.
Examples with Solutions
Worked Examples: Repeated Roots & Initial Value Problems
Example 1
Solve:
\[ \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + y = 0 \]Solution:
The auxiliary equation:
\[ k^2 + 2 k + 1 = 0 \implies (k + 1)^2 = 0 \implies k = -1 \]The general solution:
\[ y(x) = A e^{-x} + B x e^{-x} \]where \( A \) and \( B \) are constants. You can verify by substitution that this satisfies the differential equation.
Example 2
Solve with initial conditions \( y(0) = 4 \) and \( y'(0) = 0 \):
\[ \frac{d^2 y}{dx^2} - 4 \frac{dy}{dx} + 4 y = 0 \]Solution:
The auxiliary equation:
\[ k^2 - 4 k + 4 = 0 \implies (k - 2)^2 = 0 \implies k = 2 \]The general solution:
\[ y(x) = A e^{2x} + B x e^{2x} \]Apply initial conditions:
\[ y(0) = A = 4 \]Differentiate to find \( y'(x) \):
\[ y'(x) = 2 A e^{2x} + B (e^{2x} + 2 x e^{2x}) \]Using \( y'(0) = 0 \):
\[ y'(0) = 2A + B = 0 \implies 2(4) + B = 0 \implies B = -8 \]Final solution:
\[ y(x) = 4 e^{2x} - 8 x e^{2x} \]