× Close
Join the Ques10 Community
Ques10 is a community of thousands of students, teachers, and academic experts, just like you.
Join them; it only takes a minute
Sign up
Question: Solve $(D^3+D)y=cosec(x)$ by the method of variation of parameters.
0

Subject : Applied Mathematics 2

Topic : Linear differential equation with constant coefficients

Difficulty : High

am2(82) • 232 views
ADD COMMENTlink
modified 5 months ago by gravatar for awari.swati831 awari.swati831140 written 7 months ago by gravatar for smitapn612 smitapn6120
0

$ \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 \\ $

ADD COMMENTlink
modified 5 months ago by gravatar for awari.swati831 awari.swati831140 written 6 months ago by gravatar for Saurabh  Singh Saurabh Singh0
Please log in to add an answer.


Use of this site constitutes acceptance of our User Agreement and Privacy Policy.