Answer to Question #252035 in Differential Equations for Shagor

Question

Apply the method of Varition of parameter

d^2y/dx^2 -4 dy/dx +3 y = 1/1+e^-x

Answer 1

$\frac{d^2y}{dx^2}-4\frac{dy}{dx}+3y=\frac{1}{1+e^{-x}}\\ y''-4y'+3=\frac{1}{1+e^{-x}}\\ \text{The auxilliary equation is; }\\ m^2-4m+3=0\\ (m-1)(m-3)=0\\ m=1 \text{ or } 3\\ \text{The complimentary solution is; }\\ y=Ae^x+Be^{3x}\\ \text{The Wronskian for the DE is;}\\ W=y_1y_2'-y_1'y_2\\ W=3e^{4x}-e^4x=2e^4x\\ C_1'=\frac{-f(x)y_2}{W}=\frac{e^{3x}}{2e^{4x}(1+e^{-x})}=\frac{1}{2e^x(1+e^{-x})}=\frac{1}{2e^x+2}\\ C_2'=\frac{f(x)y_1}{W}=\frac{e^x}{2e^{4x}(1+e^{-x})}=\frac{1}{(2e^{3x}+2e^{2x})}\\ C_1=\int \frac{1}{2e^x(1+e^{-x})} dx=\frac{1}{2}(x-\ln(e^x+1))+K_1\\ C_2=\int \frac{1}{(2e^{3x}+2e^{2x})} dx= \frac{1}{4}(2x-e^{-2x}+2e^{-x}-2\ln(e^x+1))+K_2\\ \text{Therefore, }\\ y(x)=Ae^x+Be^{3x}+\frac{e^x}{2}(x-\ln(e^x+1))+\frac{e^{3x}}{4}(2x-e^{-2x}+2e^{-x}-2\ln(e^x+1))+K$