Answer to Question #315651 in Differential Equations for Sheshank

Question

1. Solve the ordinary linear differential equation by the method of undetermined coefficients: y'' -2y' -3y=2e^ 4x

Answer 1

Solution

For the homogeneous equation y'' - 2y' - 3y = 0 the characteristic equation is

λ² - 2λ - 3=0  => $\lambda_{1,2}=1\pm\sqrt{1+3}$   => λ₁ = -1, λ₂ = 3

So the solution of homogeneous equation is y₀(x) = C₁e^-x+C₂e^3x, where C₁, C₂ are arbitrary constants.

Partial solution may be find in the form y₁(x) = Ae^4x. Substitution into equation y'' -2y' -3y=2e^4x gives

(16Ae^4x -2*4A-3A) e^4x  = 2e^4x => 5A = 2 => A = 0.4   

Therefore the solution of given equation is

y(x) = y1(x)+y2(x) = C₁e^-x+C₂e^3x +0.4 e^4x