Question

What is the general solution of the differential equation `(d^2y)/(dx^2) + 3dy/dx + 2y = 2sinx`?

Answer 1

`y(x) = Ae^(-x)+Be^(-2x) -3/5cosx+1/5sinx`

Explanation:

We have:

`(d^2y)/(dx^2) + 3dy/dx + 2y = 2sinx` ..... [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''+3y'+2y = 0`

And it's associated Auxiliary equation is:

`m^2+3m+2 = 0`
`(m+1)(m+2) = 0`

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

Thus the solution of the homogeneous equation is:

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

Particular Solution

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

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

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

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

`-acos(x)-bsin(x) + 3{-asin(x)+bcos(x)} + 2{acos(x)+bsin(x)}=2sin(x)`

`-acos(x)-bsin(x) -3asin(x)+3bcos(x) + 2acos(x)+2bsin(x)=2sin(x)`

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

`cos(x): -a+3b+2a=0 => a+3b=0`
`sin(x): -b -3a+2b=2 \ => b-3a=2`

Solving simultaneously we get:

`a=-3/5` and `b=1/5`

And so we form the Particular solution:

`y_p = -3/5cosx+1/5sinx`

General Solution

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Ae^(-x)+Be^(-2x) -3/5cosx+1/5sinx`