Use reduction of order to find a second linearly independent solution (2x+5)y′′−4(x+3)y′+4y=0,x>−52,y1 =e2x

Given that exp(2x) is a solution, we assume another solution of the form

y(x) = v(x) exp(2x) = v exp(2x)

with derivatives

y' = v' exp(2x) + 2v exp(2x)

y'' = v'' exp(2x) + 4v' exp(2x) + 4v exp(2x)

Substitute these into the equation:

(2x + 5) (v'' exp(2x) + 4v' exp(2x) + 4v exp(2x)) - 4 (x + 3) (v' exp(2x) + 2v exp(2x)) + 4v exp(2x) = 0

Each term contains a factor of exp(2x) that can be divided out:

(2x + 5) (v'' + 4v' + 4v) - 4 (x + 3) (v' + 2v) + 4v = 0

Expanding and simplifying eliminates the v term:

(2x + 5) v'' + (4x + 8) v' = 0

Substitute w(x) = v'(x) to reduce the order of the equation, and you're left with a linear ODE:

(2x + 5) w' + (4x + 8) w = 0

w' + (4x + 8)/(2x + 5) w = 0

I'll use the integrating factor method. The IF is

µ(x) = exp( ∫ (4x + 8)/(2x + 5) dx ) = exp(2x - log|2x + 5|) = exp(2x)/(2x + 5)

Multiply through the ODE in w by µ :

µw' + µ (4x + 8)/(2x + 5) w = 0

The left side is the derivative of a product:

[µw]' = 0

Integrate both sides:

∫ [µw]' dx = ∫ 0 dx

µw = C

Replace w with v', then integrate to solve for v :

exp(2x)/(2x + 5) v' = C

v' = C (2x + 5) exp(-2x)

∫ v' dx = ∫ C (2x + 5) exp(-2x) dx

v = C₁ (x + 3) exp(-2x) + C₂

Replace v with y exp(-2x) and solve for y :

y exp(-2x) = C₁ (x + 3) exp(-2x) + C₂

y = C₁ (x + 3) + C₂ exp(2x)

It follows that the second fundamental solution is y = x + 3. (The exp(2x) here is already accounted for as the first solution.)

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