Find the general solution of
y′′−2y′−3y=4e3x+ex(cos(x)−2sin(x))\displaystyle {y}{''}-{2}{y}'-{3}{y}={4}{e}^{{{3}{x}}}+{e}^{{x}}{\left({\cos{{\left({x}\right)}}}-{2}{\sin{{\left({x}\right)}}}\right)}y′′−2y′−3y=4e3x+ex(cos(x)−2sin(x))
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