Xd^2y/dx^2-4xdy/dx+4xy=5xe^(3x)+4xe^(2x)??

1 Answer
Jul 6, 2018

Answer:

#y = e^(2x)( 5e^x+ 2x^2 + c_1 x + c_2 )#

Explanation:

#x(d^2y)/dx^2-4xdy/dx+4xy=5xe^(3x)+4xe^(2x)#

Eliminate common #x# term

#(d^2y)/dx^2-4dy/dx+4y=5e^(3x)+4e^(2x)#

Expressing in terms of linear differential operator #D = d/(dx)#:

#(D^2-4D+4)y=5e^(3x)+4e^(2x)#

Factoring:

#(D-2)underbrace((D-2)y)_(= z)=5e^(3x)+4e^(2x) qquad triangle#

#z' - 2 z=5e^(3x)+4e^(2x)#

Integrating factor: #exp(int (-2) dx) = e^(- 2x)#

Distribute integrating factor across equation:

# (e^(-2x) z)^' =5e^x+4 #

Integrating:

# e^(-2x) z =5e^x+4x + c_1#

# :. z =5e^(3x)+4xe^(2x) + c_1e^(2x) qquad bb(= (D-2)y)#

#y'-2 y = 5e^(3x)+4xe^(2x) + c_1e^(2x) #

Using same integrating factor again:

#(e^(-2x)y)^' = 5e^( x)+4x + c_1 #

Integrating:

# e^(-2x)y = 5e^( x)+ 2x^2 + c_1 x + c_2#

#y = 5e^( 3x)+ 2x^2e^(2x) + c_1 xe^(2x) + c_2e^(2x)#

# = e^(2x)( 5e^x+ 2x^2 + c_1 x + c_2 )#