Answer to Question #117823 in Differential Equations for Aldous

Question

Find a general solution for the differential equation

(D^2 - 3D +2) = (e^2x)/(1+e^2x)

Answer 1

the auxiliary equation f the given differential equation is:

D²-3D+2=0

D=2,1

y(x)= c₁e^2x+c₂e^x

particular solution by method of variation of parameters:

let u(x)=e^2x

v=e^x

f= $\frac{e^{2x}}{1+e^{2x}}$

then, wronskian W(u,v)=$\begin{vmatrix} e^{2x} & e^{x} \\ 2e^{2x} & e^{x} \end{vmatrix}$ = -e^3x

then, the particular solution is:

$=-u\int{\frac{v.fdx}{W}}+v\int{\frac{u.fdx}{W}}$

$= -e^{2x}\int{\frac{e^x . e^{2x}dx}{-e^{3x}(1+e^{2x})}}+ e^{x}\int{\frac{e^{2x} . e^{2x}dx}{-e^{3x}(1+e^{2x})}}$

$=e^{2x} \int{\frac{1}{1+e^{2x}}}dx - e^x\int{\frac{e^x}{1+e^{2x}}}dx$

$= e^{2x}.ln|sin(tan^{-1}.e^x)|-e^x tan^{-1}e^x$

so the solution is:

$y=c_1e^{2x}+c_2e^x+e^{2x}.ln|sin(tan^{-1}.e^x)|-e^x tan^{-1}e^x$