Show that xsin2x is a solution to the DE y'' + 4y = 4cos2x ?
2 Answers
I tried this:
Explanation:
To show this you need to derive twice your solution
Let us derive:
again:
let us substitute
that gives
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
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