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Question: Solve $(D^3+D)y=cosec(x)$ by the method of variation of parameters.
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Subject : Applied Mathematics 2

Topic : Linear differential equation with constant coefficients

Difficulty : High

am2(82) • 58 views
 modified 9 days ago by written 6 weeks ago by
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$\text{ The Auxillary equation is } D^3 + D = 0 \Longrightarrow D = 0, \pm i \\$

$\text{ C.F. is } y_c = c_1 + c_2cosx + c_3sinx \\$

$\text {Let }y_p = u + vcosx + wsinx = uy_1 + vy_2 + wy_3 \\$

$W = \begin{vmatrix} y_1 & y_2 & y_3 \\ {y'_1} & {y'_2} & {y'_3} \\ {y''_1} &{y''_2}&{y''_3} \end{vmatrix} \\$

$= \begin{vmatrix} 1 & cosax & sinax \\ 0 & -asinax & acosax \\ 0 & -cosx & -sinx \end{vmatrix} \\$

$= sin^2x + cos^2x = 1 \\$

$u = \int \frac{y_2 y'_3 -y_3y'_2} {W} x dx = \int \frac{ [cosx (-cosx) - sinx (-sinx)]}{1} cosecx dx \\$

$= \int cosecxdx = log(cosecx-cotx) \\$

$v = \int \frac{y_3 y'_1 -y_1 y'_3} {W} x dx = \int \frac{ [sinx (0) - 1(cosx)]}{1} cosecx dx \\$

$= - \int cotxdx = - log(sinx) \\$

$w = - \int \frac{y_1 y'_2 -y_2 y'_1} {W} x dx = \int \frac{ [1 (-sinx) - cosx (0)]}{1} cosecx dx \\$

$= -x \\$

$\text{P.I = } y_p = log(cosecx-cotx) + (- log(sinx)) -xsinx \\$

$\therefore \text{ The complete solution is y = C.F. + P.I. } \\$

$\therefore y = c_1 + c_2cosx + c_3sinx + log(cosecx-cotx) - log(sinx) -xsinx \\$