Question

(nonhomogeneous linear ODEs) `y''+4y'=-12sin2x, (y_h+y_p=)`?

Answer 1

`y(x) = A + Be^(-4x) + 6/5cos2x+3/5sin2x`

Explanation:

We have:

`y''+4y'=-12sin2x` .... [A}

This is a second order linear non-Homogeneous Differentiation Equation with constant coefficients. 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''+4y = 0`

And it's associated Auxiliary equation is:

`m^2+4m = 0 => m(m+4)`

Which has two real and distinct solution `m=0,-4`

Thus the solution of the homogeneous equation is:

`y_c = Ae^(0x) +_Be^(-4x)`
`\ \ \ = A + Be^(-4x)`

Particular Solution

In order to find a particular solution of the non-homogeneous equation we would look for a solution of the form:

`y = (acos2x+bsin2x)`

Where the constants `a` and `b` are to be determined by direct substitution and comparison:

Differentiating wrt `x` we get:

`dy/dx = -2asin2x+2bcos2x`

Differentiating again wrt `x` we get:

`(d^2y)/(dx^2) = -4acos2x+4bsin2x`

Substituting into the DE [A] we get:

`(-4acos2x+4bsin2x) + 4(-2asin2x+2bcos2x) = -12sin2x`
`:. -4acos2x+4bsin2x + -8asin2x+8bcos2x = -12sin2x`
`:. (-4a+8b)cos2x+(4b-8a) = -12sin2x`

Equating coefficients of `sin2x` and `cos2x` we get

`cos2x: -4a+8b = 0`
`sin2x: -4b-8a=-12`

Solving simultaneously, we have:

`a = 6/5`
`b = 3/5`

And so we form the Particular solution:

`y_p = 6/5cos2x+3/5sin2x`

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = A + Be^(-4x) + 6/5cos2x+3/5sin2x`

Answer 2

See below.

Explanation:

For the homogeneous solution

`y''_h+4y'_h=0` Making `y_h = e^(lambdax)` we have

`e^(lambda x) lambda (lambda+4)=0` and then

`y_h=c_1e^0+c_2e^(-4x) = c_1+c_2e^(-4x)`

for the particular solution `y_p` we substitute

`y_p=a sin(2x)+b cos(2x)` into the complete equation thus obtaining

`(b-2a) Cos(2 x) + (a + 2 b-3) Sin(2 x)=0`

this must be true for all `x` then

`{(b-2a=0),(a + 2 b-3=0):}` and solving

`a=3/5` and `b = 6/5` and finally

`y=y_h+y_p = c_1+c_2e^(-4x)+3/5 sin(2x)+6/5 cos(2x)`