Answer to Question #225913 in Chemical Engineering for Lokika

Question

Solution of (D³ - 2D² - 3D)y = 0

A) y = C1.₁e^{x} + C₂е^{-x }+ С3.e^{3x} B) y = C1e^x + C₂е^{-x} + С3.e^{-3x}

C) y = C1+C₂е^{-x} + С3.e^{3x}

D) y = C₁+C₂е^{-x} + С3e^{-3x}

Answer 1

$\left(D^3-2D^2-3D\right)y=0\\ \mathrm{A\:linear\:homogeneous\:ODE\:with\:constant\:coefficients\:has\:the\:form\:of}\:\left(a_nD^n+...+a_1D+a_0\right)y=0\\ D=0,\:D=-1,\:D=3\\ \mathrm{For\:non\:repeated\:real\:roots\:}D_1,\:D_2,\:...,\:D_n\mathrm{,\:the\:general\:solution\:takes\:the\:form:}\\ y=c_1e^{D_1\:x}+c_2e^{D_2\:x}+...+c_ne^{D_n\:x}\\ c_1e^0+c_2e^{-x}+c_3e^{3x}\\ y=c_1+c_2e^{-x}+c_3e^{3x}$