Answer to Question #122607 in Differential Equations for jessica

Question

1. The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2,

y2 = y1(x) ∫ e^−∫P(x) dx / y upper 2 and lower 1 (x) dx

as instructed, to find a second solution y2(x).
y'' − 4y' + 4y = 0; y1 = e^2x

y2 = _____

Answer 1

Solution:

y₁(x)=e^2x,

P(x)=-4,

$\int$ ^x₀P(x')dx'=$\int$^x₀ (-4)dx'=-4x'|^x₀=-4x,

y₂(x)=y₁(x)$\int$e^(- $\int$ ^x₀ (-4)dx')/y₁²(x)dx=e^2x$\int$ e^4x/e^4xdx=e^2x$\int$ dx=e^2x(x+C),

Let C=0, then y₂(x)=xe^2x

Answer: y₂(x)=xe^2x.