Question

Show that `xsin2x` is a solution to the DE `y'' + 4y = 4cos2x`?

Answer 1

I tried this:

Explanation:

To show this you need to derive twice your solution `y=xsin(2x)` and substitute for `Y' '` and `y` into your equation and see if it works.
Let us derive:
`Y'=sin(2x)+2xcos(2x)`
again:
`Y' '=2cos(2x)+2cos(2x)-4xsin(2x)`

let us substitute `Y' '` and the given result `y=xsin(2x)` into our equation#:

`2cos(2x)+2cos(2x)-4xsin(2x)+4(xsin(2x))=4cos(2x)`

that gives `0=0` implying that the function `y=xsin(2x)` is solution of our equation.

Answer 2

We seek to show that:

`y=xsin2x`

is a solution to the DE:

`y'' + 4y = 4cos2x` ..... [A]

Given that we have the solution we can just differentiate (twice) and substitute:

`y' = (x)(d/dxsin2x)+(d/dxx)(sin2x)`
`\ \ \ \ = 2xcos2x+sin2x`

`y'' = (2x)(d/dxcos2x) + (d/dx2x)(cos2x) + 2cos2x`
`\ \ \ \ \ = -4xsin2x + 2cos2x + 2cos2x`
`\ \ \ \ \ = -4xsin2x + 4cos2x`

Substituting into the DE [A], we get:

`y'' + 4y = (-4xsin2x + 4cos2x) + 4(xsin2x)`
`" " = -4xsin2x + 4cos2x + 4xsin2x`
`" " = 4cos2x` QED.

Note: this solution, is known as the Particular Solution. We could go on and form the General Solution as follows:

Complementary Function

The homogeneous equation associated with [A] is

`y'' + 4y = 0` ..... [B]

And it's associated Auxiliary equation is:

`m^2 + 4 = 0`

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

Thus the solution of the homogeneous equation [B] is:

`y_c = e^0(Acos(2x) + Bsin(2x))`
`\ \ \ = Acos2x + Bsin2x`

Which then leads to the GS of [A}

`y(x) = y_c + y_p =Acos2x + Bsin2x` `+ xsin2x`