Answer to Question #220815 in Differential Equations for Unknown346307

Question

Solve (x + 2)y"-(2x + 5)y'

+ 2y = (x + 1)ex

Answer 1

Solution

One of the solutions of homogeneous equation (x + 2)y₀"-(2x + 5)y₀'+ 2y₀ = 0 let’s find in the form y₀(x) = e^rx. Substitution into equation gives

(x + 2)r²-(2x + 5)r+ 2= 0   =>   x(r²-2r)+(2r²-5r+2) = 0  => from first brackets equal to zero r=0, r=2 and from second r = 1/2, r = 2  =>  r = 2

So y₀₁(x) = e^2x.

Once we have this first solution we will then assume that a second solution of homogeneous equation will have the form

y(x) = y₀₁(x)u(x)

Plugging this into the homogeneous equation gives

(x+2)u’’+(2x+3)u’ = 0

New function w(x) = u’(x)  => (x+2)w’ = -(2x+3)w  =>  dw/w = -(2x+3)dx/(x+2)  => ln|w| = - 2x+ln|x+2| + C₁  =>  w = C₁ (x+2)e^-2x  => u(x) = C₂+∫w(x)dx = C₂+ C₁∫(x+2)e^-2x dx = C₂ - C₁ (x/2+5/4) e^-2x    =>   y₀(x) = y₀₁(x)u(x) = C₂ e^2x - C₁ (x/2+5/4)    

Partial solution of initial nonhomogeneous equation let’s find in the form y₁(x) = Ae^x  

Plugging this into the nonhomogeneous equation gives

A[(x+2)-(2x-5)+2] = x+1  => A(-x-1) = x+1  = >   A = -1  =>  y₁(x) = - e^x    

Therefore solution of initial equation is

y(x) = y₀(x) + y₁(x) = C₂ e^2x - C₁ (x/2+5/4) - e^x