Answer to Question #221447 in Differential Equations for Divya

Question

By method variation of parameters solve the ODE(D^2+4)y=2 cosec2x?

Answer 1

Answer:-

General solution of $y''+4y=0$ is $y=A\cos 2x+B\sin 2x$.

Let $y=A(x)\cos 2x+B(x)\sin 2x$ and $A'(x)\cos 2x+B'(x)\sin2x=0$. Then $y'=-2A(x)\sin 2x+2B(x)\cos 2x$ and so $y''=-2A'(x)\sin 2x+2B'(x)\cos 2x-4A(x)\cos 2x-4B(x)\sin 2x$.

We have $\frac{1}{2\sin 2x}=y''+4y=-2A'(x)\sin 2x+2B'(x)\cos 2x$. Also we have $A'(x)\cos 2x+B'(x)\sin2x=0$.

$-1=0\cdot 2\cos 2x-\frac{1}{2\sin 2x}\sin 2x=$

$=2(A'(x)\cos 2x+B'(x)2\sin2x)\cos 2x-$

$-(-2A'(x)\sin 2x+2B'(x)\cos 2x)\sin 2x=2A'(x)$, so $A'(x)=-\frac{1}{2}$ and $A(x)=-\frac{1}{2}x+A$

Since $0=A'(x)\cos 2x+B'(x)\sin2x=-\frac{1}{2}\cos 2x+B'(x)\sin2x$, we have $B'(x)=\frac{1}{2}\frac{\cos 2x}{\sin 2x}$. Then $dB=\frac{1}{2}\frac{\cos 2x}{\sin 2x}=\frac{1}{4}\frac{d\sin 2x}{\sin 2x}=\frac{1}{4}d\ln|\sin 2x|$ and $B(x)=\frac{1}{4}\ln|\sin 2x|+B$

We obtain $y=\left(-\frac{1}{2}x+A\right)\cos 2x+\left(\frac{1}{4}\ln|\sin 2x|+B\right)\sin 2x$

Answer: $y=\left(-\frac{1}{2}x+A\right)\cos 2x+\left(\frac{1}{4}\ln|\sin 2x|+B\right)\sin 2x$