Question #151931

Find a linear homogeneous differential equation, whose solutions are {e^x,e^-x,e^x-2e^-x}

Expert's answer

Let us solve the linear homogeneous differential equation yy=0y''-y=0. The characteristic equation k21=0k^2-1=0 has the solutions k1=1k_1=1 and k2=1k_2=-1. Therefore, the general solution is y=C1ex+C2exy=C_1e^x+C_2e^{-x}. If C1=1C_1=1 and C2=0C_2=0 we have y=exy=e^x. If C1=0C_1=0 and C2=1C_2=1 we have we have y=exy=e^{-x}. If C1=1C_1=1 and C2=2C_2=-2 we have y=ex2ex.y=e^x-2e^{-x}.


Answer: yy=0y''-y=0


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