Question

by using method of variation of parameter find p(D^2+1)y-cosec x​

Answer 1

Step-by-step explanation:

I am in 4std....Hiiiiiiii II

Answer 2

Solve by using method of variation of parameters

$\tt \:  ( {D}^{2} + 1)y = cosecx$

Answer :-

Given equation is

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

Step 1 :- To find general solution:-

$\tt \:  \longrightarrow \: The \: characteristic \: equation \: is \: {m}^{2} + 1 = 0$

$\tt\implies \: {m}^{2} = - 1$

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

$\tt \:  \longrightarrow \: So \: the \: general \: solution \: is \: given \: by \:$

$\tt \:  \longrightarrow \: complementary \: function \: = a \: cosx \: + \: b \: sinx - - (i)$

Step 2 :- To find the Wronskian (W) :-

Now, here

$\tt \:  \longrightarrow \: y_1 = cosx \\ \tt \:  \longrightarrow \: y_2 = sinx \\ \tt \:  \longrightarrow \: y_1' = - sinx \\ \tt \:  \longrightarrow \:y_2' = cosx$

$\tt \:  \longrightarrow \: Now, \: W \: = y_1y_2' - y_2y_1'$

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

Step 3 :- To find the particular solution by using formula :-

$\tt \:  y_p(x) = - y_1 \int \: \dfrac{y_2 \: f(x)}{W} dx + y_2 \int \: \dfrac{y_1 \: f(x)}{W} dx$

Step 4 :- Evaluation of integrals :-

$\tt \:  \longrightarrow \: Now, \: consider : \: - \: y_1 \int \: \dfrac{y_2 \: f(x)}{W}$

☆ On substituting the values we get,

$\tt \:  \longrightarrow \: = \: - cosx \int \: \dfrac{( sin \: x)\: cosecx}{1} dx$

$\tt \:  \longrightarrow \: = - \: cosx \int \: dx$

$\tt \:  \longrightarrow \: = - \: x \: cosx$

$\tt \:  \longrightarrow \: Now, \: consider : \: \: y_2 \int \: \dfrac{y_1 \: f(x)}{W} dx$

$\tt \:  \longrightarrow \: = sin \: x \: \int \: \dfrac{cosx \: cosecx}{ 1} dx$

$\tt \:  \longrightarrow \: = sinx \: \int \: cotx \: dx$

$\tt \:  \longrightarrow \: = sinx \: log \: (sinx)$

Step 5 :- Complete solution is given by

$\tt \:  y \: = \: acosx \: + \: bsinx \: - xcosx \: + sinx \: log(sinx)$