Question

What is the general solution of the differential equation? : `(d^2y)/dx^2-dy/dx-2y=4x^2`

Answer 1

`y = Ae^(-x)+Be^(2x) -2x^2+2x-3`

Explanation:

There are two major steps to solving Second Order DE's of this form:

`(d^2y)/dx^2-dy/dx-2y=4x^2`

1) Find the Complementary Function (CF)
This means find the general solution of the Homogeneous Equation

`(d^2y)/dx^2-dy/dx-2y=0`

To do this we look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

`m^2-m-2=0`
`(m-2)(m+1) = 0`

This has two distinct real solutions, `m=-1,2`

And so the solution to the DE is;

`y = Ae^(-x)+Be^(2x)` Where `A,B` are arbitrary constants

-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Verification:

If `y = Ae^(-x)+Be^(2x)`, then
`y' = -Ae^(-x)+2Be^(2x)`
`y'' = Ae^(-x)+4Be^(2x)`

And so, `y''-y'-2y = Ae^(-x)+4Be^(2x) -(-Ae^(-x)+2Be^(2x))-2(Ae^(-x)+Be^(2x)) = 0`
-+-+-+-+-+-+-+-+-+-+-+-+-+-+

2) Find a Particular Integral* (PI)

This means we need to find a specific solution (that is not already part of the solution to the Homogeneous Equation). As the RHS is a quadratic we try a solution of the quadratic form:

`y=ax^2+bx+c`

Where `a,b,c` are constants to be found

If `y=ax^2+bx+c` then we have:

`y'\ = 2ax+b`
`y'' = 2a`

If we substitute into the initial DE we get:

`\ \ 2a - (2ax+b) - 2(ax^2+bx+c)=4x^2`

`:. 2a - 2ax-b - 2ax^2-2bx-2c = 4x^2`

Equating Coefficients we have

`Coef(x^2) : -2a=4 \ \ \ \ \ \ \ \ => a=-2`
`Coef(x^1) : -2a-2b=0 => b=2`
`Coef(x^0) : 2a-b-2c=0 \ \ \ \ => c=-3`

So we have found that a Particular Solution is:

`y = -2x^2+2x-3`

3) General Solution (GS)
The General Solution to the DE is then:

GS = CF + PI

Hence The General Solution to the initial DE is

`y = Ae^(-x)+Be^(2x) -2x^2+2x-3`