Question #117823
Find a general solution for the differential equation

(D^2 - 3D +2) = (e^2x)/(1+e^2x)
1
Expert's answer
2020-05-27T13:03:49-0400

the auxiliary equation f the given differential equation is:

D2-3D+2=0

D=2,1

y(x)= c1e2x+c2ex


particular solution by method of variation of parameters:


let u(x)=e2x

v=ex

f= e2x1+e2x\frac{e^{2x}}{1+e^{2x}}


then, wronskian W(u,v)=e2xex2e2xex\begin{vmatrix} e^{2x} & e^{x} \\ 2e^{2x} & e^{x} \end{vmatrix} = -e3x


then, the particular solution is:

=uv.fdxW+vu.fdxW=-u\int{\frac{v.fdx}{W}}+v\int{\frac{u.fdx}{W}}


=e2xex.e2xdxe3x(1+e2x)+exe2x.e2xdxe3x(1+e2x)= -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})}}


=e2x11+e2xdxexex1+e2xdx=e^{2x} \int{\frac{1}{1+e^{2x}}}dx - e^x\int{\frac{e^x}{1+e^{2x}}}dx


=e2x.lnsin(tan1.ex)extan1ex= e^{2x}.ln|sin(tan^{-1}.e^x)|-e^x tan^{-1}e^x


so the solution is:

y=c1e2x+c2ex+e2x.lnsin(tan1.ex)extan1exy=c_1e^{2x}+c_2e^x+e^{2x}.ln|sin(tan^{-1}.e^x)|-e^x tan^{-1}e^x


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