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Solve $(D^2+a^2)y=sec(ax)$ by the method of variation of parameters.

Subject : Applied Mathematics 2

Topic : Linear differential equation with constant coefficients

Difficulty : High

1 Answer
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$ \text{ The Auxillary equation is } \\ $

$ D^2 + a^2 =0 \\ $

$ D = \pm ai \\ $

$ \text{ C.F. is } y_c = c_1cosax + c_2sinax \\ $

$ \text {Let }y_p = ucosax + vsinax = uy_1 + vy_2 \\ $

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

$ = \begin{vmatrix} cosax & sinax \\ -asinax & acosax \end{vmatrix} \\ $

$ = a \\ $

$ u = - \int \frac{y_2X} {W} = \frac{-1}{a} \int [sinax (secax)] dx \\ $

$ = \frac{-1}{a} \int tanax dx \\ $

$ = \frac{-1}{a^2} log secax \\ $

$ v = \int \frac{y_1X} {W} = \frac{1}{a} \int cosaxsecax dx \\ $

$ = \frac{x}{a} \\ $

$ \text{P.I = } y_p =\frac{-1}{a^2} log secax(cosax) + \frac{x}{a}sinax \\ $

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

$ \therefore y = c_1cosax + c_2sinax - \frac{1}{a^2} log secax(cosax) + \frac{x}{a}sinax \\ $

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