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

2 Answers
Aug 18, 2017

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.

Aug 18, 2017

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