Prove that y(x) = C1 sin2x + C2 cos2x is a solution of y
(2) + 4y = 0
Given y(x)=c1sin2x+c2cos2xy′(x)=2c1cos2x−2c2sin2xy′′(x)=−4c1cos2x−4c2cos2x ⟹ y′′(x)+4y(x)=−4c1cos2x−4c2cos2x+4c1sin2x+4c2cos2x=−4c1cos2x+4c1cos2x−4c1sin2x+4c1cos2x=0∴y(x)=c1sin2x+c2cos2x is a solution of y′′(x)+4y(x)=0\text{Given } y(x) = c_1 \sin 2x + c_2 \cos 2x \\ y'(x) = 2c_1 \cos 2x - 2c_2 \sin 2x \\ y''(x)= -4c_1 \cos 2x - 4c_2 \cos 2x \\ \implies y''(x) + 4y(x) = -4c_1 \cos 2x - 4c_2 \cos 2x + 4c_1 \sin 2x + 4c_2 \cos 2x \\ \qquad \qquad \qquad \qquad \quad = -4c_1 \cos 2x + 4c_1 \cos 2x -4c_1 \sin 2x + 4c_1 \cos 2x\\ \qquad \qquad \qquad \qquad \quad = 0 \\ \therefore y(x) = c_1 \sin 2x + c_2 \cos 2x \text{ is a solution of } y''(x) + 4y(x) = 0Given y(x)=c1sin2x+c2cos2xy′(x)=2c1cos2x−2c2sin2xy′′(x)=−4c1cos2x−4c2cos2x⟹y′′(x)+4y(x)=−4c1cos2x−4c2cos2x+4c1sin2x+4c2cos2x=−4c1cos2x+4c1cos2x−4c1sin2x+4c1cos2x=0∴y(x)=c1sin2x+c2cos2x is a solution of y′′(x)+4y(x)=0
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