Question

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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this should be solution

Answer 1

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`