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.
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: \[ y(x) = A e^{k x} + B x e^{k x} \] where \(A\) and \(B\) are constants determined by initial conditions.
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.
Solve:
\[ \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + y = 0 \] Solution:The auxiliary equation:
\[ k^2 + 2 k + 1 = 0 \] \[ (k + 1)^2 = 0 \quad \Rightarrow \quad k = -1 \]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.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:
\[ k^2 - 4 k + 4 = 0 \] \[ (k - 2)^2 = 0 \quad \Rightarrow \quad k = 2 \]The general solution:
\[ y(x) = A e^{2x} + B x e^{2x} \]Apply initial conditions:
\[ y(0) = A = 4 \] \[ y'(x) = 2 A e^{2x} + B (e^{2x} + 2 x e^{2x}) \quad \Rightarrow \quad y'(0) = 2A + B = 0 \] \[ B = -2A = -8 \]Final solution:
\[ y(x) = 4 e^{2x} - 8 x e^{2x} \]