Question

Find the general solution of y"-3y'+2y=1/(1+e^(-x)), given that y1=e^2x satisfies the complementary equation.

Answer 1

y=ue^(2x)
y'=u'e^(2x)+2ue^(2x)
y"=u"e^(2x)+4u'e^(2x)+4ue^(2x)
Subbing:
y"-3y'+2y=u"e^(2x)+u'e^(2x)=1/(1+e^(-x))
\[u''+u'=\frac{ e ^{-2x} }{ 1+e ^{-x} }\]

Answer 2

z=u' subbing:
\[z'+z=\frac{ e ^{-2x} }{ 1+e ^{-x} }\]

Answer 3

the integrating factor is e^x
\[e ^{x}(z'+z)=\frac{ e ^{-x} }{ 1+e ^{-x} }\]

Answer 4

\[\frac{ d }{ dx }(e ^{x}z)=\frac{ e ^{-x} }{ 1+e ^{-x} }\]
\[e ^{x}z=\int\limits_{}^{}\frac{ e ^{-x} }{ 1+e ^{-x} }dx\]
\[e ^{x}z=-\ln \left| 1+e ^{-x} \right|+C\]
\[z=u'=-e ^{-x}\ln \left| 1+e ^{-x} \right|+Ce ^{-x}\]
\[u=(1+e ^{-x})\ln \left| 1+e ^{-x} \right|-(1+e ^{-x})+C _{2}e ^{-x}+C _{1}\]
Since y=ue^(2x),
\[y=(e ^{2x}+e ^{x})[\ln \left| 1+e ^{-x} \right|-1]+C _{1}e ^{2x}+C _{2}e ^{x}\]

Answer 5

But the answer in the book says \[y=(e ^{2x}+e ^{x})\ln(1+e ^{-x})+C _{1}e ^{2x}+C _{2}e ^{x}\]

Answer 6

@Kainui @freckles @pooja195

Answer 7

your answer is the same
\[-e^{2x}-e^{x}+C_1 e^{2x}+C_2 e^x \\ =(C_1-1)e^{2x}+(C_2-1)e^{x} \\ =K_1 e^{2x}+K_2 e^{x}\]

Answer 8

So my answer and the book's answer are both correct?

Answer 9

yep

Answer 10

Okay, then. Thank you for verifying!