Question

What is the general solution of the differential equation? : `(d^2y)/(dx^2) + dy/dx - 2y = -6sin2x-18cos2x`

Answer 1

`y(x) = -e^(-2x) + 3cos(2x)`

Explanation:

We have:

`(d^2y)/(dx^2) + dy/dx - 2y = -6sin2x-18cos2x` ..... [A]

This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, `y_c` of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, `y_p` of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

`y'' + y' - 2y = 0`

And it's associated Auxiliary equation is:

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

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

Thus the solution of the homogeneous equation is:

`y_c = Ae^(-2x)+Be^(1x)`
`\ \ \ = Ae^(-2x)+Be^(x)`

Particular Solution

With this particular equation [A], a probably solution is of the form:

`y = acos(2x)+bsin(2x)`

Where `a` and `b` are constants to be determined by substitution

Let us assume the above solution works, in which case be differentiating wrt `x` we have:

`y' \ \= -2asin(2x)+2bcos(2x)`
`y'' = -4acos(2x)-4bsin(2x)`

Substituting into the initial Differential Equation `[A]` we get:

`-4acos(2x)-4bsin(2x) -2asin(2x)+2bcos(2x) - 2acos(2x) - 2bsin(2x) = -6sin2x-18cos2x`

Equating coefficients of `cos(2x)` and `sin(2x)` we get:

`cos(2x): -4a+2b-2a=-18`
`sin(2x): -4b-2a-2b=-6`

Solving simultaneously we get:

`a=3` and `b=0`

And so we form the Particular solution:

`y_p = 3cos(2x)`

General Solution

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Ae^(-2x)+Be^(x) + 3cos(2x)` ..... [B]

Initial Conditions

We are given the initial conditions:

`y(0)=2, y'(0)=2`

Putting `x=0` in [B] we get:

`2 = Ae^0+Be^0 + 3cos0 => A+B = -1`

Differentiating [B] wrt `x` we get:

`y'(x) = -2Ae^(-2x)+Be^(x) -6 sin(2x)`

Putting `x=0` we get:

`2 = -2Ae^0+Be^0 -6 sin0 => -2A + B = 2`

Solving these two new equations simultaneously we get:

`A=-1` and `B=0`

leading to the specific solution:

`y(x) = -e^(-2x) + 3cos(2x)`