Subject : Applied Mathematics 2

Topic : Linear differential equation with constant coefficients

Difficulty : High

$ \text{ The Auxiliary equation is } \\ $

$ D^2 - 1 =0 \\ $

$ D = \pm 1 \\ $

$ \text{ C.F. is } y_c = c_1e^{x} + c_2e^{-x} \\ $

$ \text {Let }y_p = ue^x + ve^{-x} = uy_1 + vy_2 \\ $

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

$ = \begin{vmatrix} e^x & e^{-x} \\ e^x & -e^{-x} \end{vmatrix} \\ $

$ = -1 -1 = -2 \\ $

$ u = - \int \frac{y_2X} {W} = \frac{-1}{-2} \int e^{-x} [ e^{-x}sin(e^{-x}) + cos(e^{-x})] dx \\ $

$ \text{Put} e^{-x} = t \\ $

$ -e^{-x}dx = dt \\ $

$ u = \frac{-1}{2} \int [ tsint + cost] dt \\ $

$ = \frac{-1}{2} [ t(-cost) + sint + sint] \\ $

$ = \frac{-1}{2} [ t(-cost) + 2sint ] \\ $

$ = \frac{1}{2} tcost + sint \\ $

$ = \frac{1}{2} e^{-x}cos(e^{-x}) - sin(e^{-x}) \\ $

$ v = \int \frac{y_1X} {W} = \frac{-1}{2} \int e^{x} [ cos(e^{-x}) + e^{-x}sin(e^{-x})] dx \\ $

$ = \frac{-1}{2} e^x cose^{-x} \\ $

$ \text{P.I = } y_p = (\frac{1}{2} e^{-x}cos(e^{-x}) - sin(e^{-x}))e^x - \frac{1}{2} e^x cos(e^{-x})e^{-x}\\ $

$ = -e^xsin(e^{-x}) \\ $

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

$ \therefore y = c_1e^{x} + c_2e^{-x} - e^xsin(e^{-x}) \\ $