Find a general solution y (x) of non-homogeneous linear equations of the 2nd order with constant coefficients and special right side. How to solve? (pictures below) Thank you a lot!

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1 Answer
Jun 21, 2018

Solve the differential equation:

y''-4y' = -12x^2+6x-4

This is a non-homogeneous linear equation with constant coefficients, so we start by solving the characteristic equation of the corresponding homogeneous equation:

y''-4y' = 0

lambda^2 -4lambda = 0

lambda(lambda-4) = 0

lambda = {(0),(4):}

So the general solution of the homogeneous equation is:

y = C_1e^(4x) +C_2

In the complete equation, the known term is in the form e^(alphax)p(x) where alpha=0 and p(x) is a polynomial of second degree.

As alpha equals one of the solutions of the characteristic equation we must look for a particular solution in the form:

bary(x) = xe^(alphax) q(x)

where q(x) is also a polynomial of second degree, whose coefficients are determined by substituting bary in the equation:

bary = x(ax^2+bx+c) = ax^3+bx^2+cx

bary' = 3ax^2 +2bx+c

bary'' = 6ax +2b

bary'' -4 bary' = 6ax +2b -12ax^2 -8bx-4c

Then:

-12ax^2 + (6a-8b)x + (2b-4c) = -12x^2+6x-4

and equating the coefficients of the same degree in x:

{(a=1),(6a-8b=6),(2b-4c=-4):}

{(a=1),(b=0),(c=1):}

So:

bary = x^3+x

and the complete solution is:

y(x) = C_1e^(4x) +C_2+ x^3+x