Question

What is the general solution of the differential equation? : `y''+4y=2sin2x`

Answer 1

The general solution is `y=c_1cos(2x)+c_2sin(2x)-1/2xcos2x`

Explanation:

`y''+4y=2sin(2x)`

This is a second order linear, non-homogenous ODE.

The general solution can be written as

`y=y_h+y_p`

Find `y_h` by solving

`y''+4y=0`

Solve the caracteristic equation

`r^2+4=0`

`r=+-2i` where `i^2=-1`

The solution is

`y_h=c_1cos(2x)+c_2sin(2x)`

Find a particular solution of the form

`y_p=a_0xsin2x+a_1xcos2x`

`y_p'=2a_0xcos2x+a_0sin2x-2a_1xsin2x+a_1cos2x`

`y_p''=-4a_0xsin2x+2a_0cos2x+2a_0cos2x-4a_1xcos2x-2a_1sin2x-2a_1sin2x`

`=-4a_0xsin2x+4a_0cos2x-4a_1xcos2x-4a_1sin2x`

Plugging those values in the ODE

`y''+4y=2sin(2x)`

`-4a_0xsin2x+4a_0cos2x-4a_1xcos2x-4a_1sin2x+4(a_0xsin2x+a_1xcos2x)=2sin2x`

`-4a_0xsin2x+4a_0cos2x-4a_1xcos2x-4a_1sin2x+4a_0xsin2x+4a_1xcos2x=2sin2x`

`4a_0cos2x-4a_1sin2x=2sin2x`

`<=>`, `{(a_0=0),(a_1=-1/2):}`

`y_p=-1/2xcos2x`

The general solution is `y=c_1cos(2x)+c_2sin(2x)-1/2xcos2x`

Answer 2

`y(x) = Acos(2x)+Bsin(2x) -1/2xcos(2x)`

Explanation:

We have:

`y'' + 4y=2sin2x` ..... [A]

This is a second order 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''+4y= 0`

And it's associated Auxiliary equation is:

`m^2+4 = 0`

Which has two pure imaginary solutions `m = +-2i`

Thus the solution of the homogeneous equation is:

`y_c = e^(0x){Acos(2x)+Bsin(2x)}`
`\ \ \ = Acos(2x)+Bsin(2x)`

Particular Solution

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

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

Where `a` and `b` are constants to be determined by substitution. However, this solution is part of `y_c`, thus we seek an alternate independant solution of the form:

`y = axcos(2x)+bxsin(2x)`

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

`y' \ \= (b-2ax)sin2x+(2bx+a)cos2x`
`y'' = -4(ax-b)cos2x-4(bx+a)sin2x`

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

`{-4(ax-b)cos2x-4(bx+a)sin2x} + 4{axcos(2x)+bxsin(2x)} = 2sin2x`

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

`cos(2x): 4b = 0 => b =0`
`sin(2x): -4a=2 => a=-1/2`

And so we form the Particular solution:

`y_p = -1/2xcos(2x)`

General Solution

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Acos(2x)+Bsin(2x) -1/2xcos(2x)`