Answer to Question #221444 in Differential Equations for Divya

Question

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

Answer 1

$\left( {{D^2} + 4} \right)y = y'' + 4y = 2{\rm{cosec}}2x$

Characteristic equation:

${k^2} + 4 = 0 \Rightarrow {k^2} = - 4 \Rightarrow k = \pm 2i$

Then the general so lution of the homogeneous equation is

${y_0} = {C_1}\cos 2x + {C_2}\sin 2x$

Let

${C_1} = f(x),\,\,{C_2} = g(x)$

Then

$y = f\cos 2x + g\sin 2x \Rightarrow y' = f'\cos 2x - 2f\sin 2x + g'\sin 2x + 2g\cos 2x$

Let

$f'\cos 2x + g'\sin 2x = 0\,\,\,\,(*)$

Then

$y' = - 2f\sin 2x + 2g\cos 2x \Rightarrow y'' = - 2f'\sin 2x - 4f\cos 2x + 2g'\cos 2x - 4g\sin 2x$

Substitute the found values ​​into the original equation:

$- 2f'\sin 2x - 4f\cos 2x + 2g'\cos 2x - 4g\sin 2x + 4f\cos 2x + 4g\sin 2x = 2{\rm{cosec}}2x$

$- 2f'\sin 2x + 2g'\cos 2x = 2{\rm{cosec}}2x$

Together with equation (*) we have the system of equations:

$\left\{ \begin{array}{l} - 2f'\sin 2x + 2g'\cos 2x = 2{\rm{cosec}}2x\\ f'\cos 2x + g'\sin 2x = 0 \end{array} \right.$

From the second equation we obtain

$g' = - f' \cdot \frac{{\cos 2x}}{{\sin 2x}}$

Substitute the found value into the first equation:

$- 2f'\sin 2x + 2\left( { - f' \cdot \frac{{\cos 2x}}{{\sin 2x}}} \right)\cos 2x = 2{\rm{cosec}}2x$

$- 2f'\sin 2x - 2f' \cdot \frac{{{{\cos }^2}2x}}{{\sin 2x}} = \frac{2}{{\sin 2x}}$

$- 2f' \cdot \frac{{{{\sin }^2}2x + {{\cos }^2}2x}}{{\sin 2x}} = \frac{2}{{\sin 2x}}$

$\frac{{ - 2f'}}{{\sin 2x}} = \frac{2}{{\sin 2x}}$

$f' = - 1 \Rightarrow f = - x + {C_3}$

$g' = - f' \cdot \frac{{\cos 2x}}{{\sin 2x}} = \frac{{\cos 2x}}{{\sin 2x}} \Rightarrow g = \int {\frac{{\cos 2x}}{{\sin 2x}}dx} = \frac{1}{2}\int {\frac{{d\sin 2x}}{{\sin 2x}} = \frac{1}{2}} \ln |\sin 2x| + {C_4}$

Then

$y = f\cos 2x + g\sin 2x = \left( { - x + {C_3}} \right)\cos 2x + \left( {\frac{1}{2}\ln |\sin 2x| + {C_4}} \right)\sin 2x$

Answer: $y =\left( { - x + {C_3}} \right)\cos 2x + \left( {\frac{1}{2}\ln |\sin 2x| + {C_4}} \right)\sin 2x$