what do i set y= to to solve this?
y'' + 4y=sin2x
ive tried y=asin2x + bcos2x and not had much luck. help please
Hello, sheep99!
What do i set y= to to solve this?
. . . $\displaystyle y'' + 4y \:=\:\sin2x$
i've tried: $\displaystyle y \:=\:A\sin2x + B\cos2x$ . . . and not had much luck.
Of course not!
$\displaystyle y \:=\:A\cos2x + B\cos2x$ is a solution of the homogeneous equation.
You must try: .$\displaystyle y \;=\;Ax\cos2x + Bx\sin2x$
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I came up with: .$\displaystyle y_p \;=\;-\tfrac{1}{4}x\cos2x$
The complete solution is: .$\displaystyle y \;=\;C_1\cos2x + C_2\sin2x - \tfrac{1}{4}x\cos2x$