Question

Solve by the method of variation of parameters y''+4 y = cosec2x​

Answer 1

Step-by-step explanation:

$General solution of y''+4y=0y′′+4y=0 is y=A\cos 2x+B\sin 2xy=Acos2x+Bsin2x.</p><p></p><p>Let y=A(x)\cos 2x+B(x)\sin 2xy=A(x)cos2x+B(x)sin2x and A'(x)\cos 2x+B'(x)\sin2x=0A′(x)cos2x+B′(x)sin2x=0. Then y'=-2A(x)\sin 2x+2B(x)\cos 2xy′=−2A(x)sin2x+2B(x)cos2x and so y''=-2A'(x)\sin 2x+2B'(x)\cos 2x-4A(x)\cos 2x-4B(x)\sin 2xy′′=−2A′(x)sin2x+2B′(x)cos2x−4A(x)cos2x−4B(x)sin2x.</p><p></p><p>We have \frac{1}{\sin 2x}=y''+4y=-2A'(x)\sin 2x+2B'(x)\cos 2xsin2x1=y′′+4y=−2A′(x)sin2x+2B′(x)cos2x. Also we have A'(x)\cos 2x+B'(x)\sin2x=0A′(x)cos2x+B′(x)sin2x=0.</p><p></p><p></p><p>-1=0\cdot 2\cos 2x-\frac{1}{\sin 2x}\sin 2x=−1=0⋅2cos2x−sin2x1sin2x=</p><p></p><p>=2(A'(x)\cos 2x+B'(x)\sin2x)\cos 2x-=2(A′(x)cos2x+B′(x)sin2x)cos2x−</p><p></p><p>-(-2A'(x)\sin 2x+2B'(x)\cos 2x)\sin 2x=2A'(x)−(−2A′(x)sin2x+2B′(x)cos2x)sin2x=2A′(x), so A'(x)=-\frac{1}{2}A′(x)=−21 and A(x)=-\frac{1}{2}x+AA(x)=−21x+A</p><p></p><p></p><p>Since 0=A'(x)\cos 2x+B'(x)\sin2x=-\frac{1}{2}\cos 2x+B'(x)\sin2x0=A′(x)cos2x+B′(x)sin2x=−21cos2x+B′(x)sin2x, we have B'(x)=\frac{1}{2}\frac{\cos 2x}{\sin 2x}B′(x)=21sin2xcos2x. Then dB=\frac{1}{2}\frac{\cos 2x}{\sin 2x}=\frac{1}{4}\frac{d\sin 2x}{\sin 2x}=\frac{1}{4}d\ln|\sin 2x|dB=21sin2xcos2x=41sin2xdsin2x=41dln∣sin2x∣ and B(x)=\frac{1}{4}\ln|\sin 2x|+BB(x)=41ln∣sin2x∣+B</p><p></p><p></p><p>We obtain y=\left(-\frac{1}{2}x+A\right)\cos 2x+\left(\frac{1}{4}\ln|\sin 2x|+B\right)\sin 2xy=(−21x+A)cos2x+(41ln∣sin2x∣+B)sin2x</p><p></p><p>Answer: y=\left(-\frac{1}{2}x+A\right)\cos 2x+\left(\frac{1}{4}\ln|\sin 2x|+B\right)\sin 2xy=(−21x+A)cos2x+(41ln∣sin2x∣+B)sin2x</p><p></p><p>$