Question

Solve by Method of Variation of parameter

y'' + y = cosecx

​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm \: y'' + y = cosecx$

Its Symbolic form is

$\rm \: ( {D}^{2} + 1)y = cosecx$

So, Auxiliary equation is

$\rm \: {m}^{2} + 1 = 0$

$\rm \: {m}^{2} = - 1$

$\rm\implies \:m \: = \: \pm \: i$

So, Complementary function is given by

$\rm \: CF = c_1cosx + c_2sinx$

So, it means

$\rm \: y_1 = cosx$

$\rm \: y_2 = sinx$

So,

$\rm \: y_1' = \: - \: sinx$

$\rm \: y_2' = cosx$

Now, Wronskian, W is given by

$\rm \: W = y_1y_2' - y_2y_1'$

$\rm \: W = {cos}^{2}x + {sin}^{2}x = 1$

Now, we know

Particular Integral is given by

$\rm \: P.I. = Py_1 + Qy_2$

where,

$\rm \: P = - \displaystyle\int\rm \frac{y_2 \: cosecx}{W} \: dx$

and

$\rm \: Q =\displaystyle\int\rm \frac{y_1 \: cosecx}{W} \: dx$

So, on substituting the values in PI, we get

$\rm \: P.I. = - cosx\displaystyle\int\rm \frac{sinx \: cosecx}{1} \: dx + sinx\displaystyle\int\rm \frac{cosx \: cosecx}{1}dx$

$\rm \: P.I. = - cosx\displaystyle\int\rm 1\: dx + sinx\displaystyle\int\rm cotxdx$

$\rm \: P.I. = - x \: cosx + sinx \: log |sinx|$

So, complete solution is

$\rm \: y = CF + P.I.$

$\rm \: y = c_1cosx + c_2sinx - x \: cosx + sinx \: log |sinx|$

Hence,

$\boxed{\rm{y = c_1cosx + c_2sinx - x \: cosx + sinx \: log |sinx| \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$