𝑦′′ − 5𝑦′ + 6𝑦 = 2𝑒^x How to solve this equation by using the method of variation of parameters?

Answer 1

Characteristic equation: r² - 5r + 6 = 0; r = 2 or r = 3.

WE have fundamental solution: y₁ = e^2x, y₂ = e^3x.

General solution of homogenius equation is y = Ae^2x+ Be^3x

Now Wronskian W(y₁, y₂) = 3e^2x·e^3x - 2e^2x·e^3x = e^5x. Now particular solution of given equation:

y_p(x) = −y₁(x)∫y₂(x)f(x)/W(y₁, y₂)dx + y₂(x)∫y₁(x)f(x)/W(y₁, y₂)dx;

y_{p(x) = - e}^2x_∫e^3x2 ·e^x/e^5xdx + e^3x∫e^2x2e^x/e^5xdx = - e^2x∫2e^-xdx + e^3x∫2e^-2xdx = 2e^2x·e^-x - e^3xe^-2x = e^x.

So, general solution of given nongomojenius ODE: y = Ae^2x + Be^3x + e^x.