Question

Find the general solution of (1-2x)y"+2y'+(2x-3)y=(1-4x+4x^2)e^x, given that y1=e^x satisfies the complementary equation.

Answer 1

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

Answer 2

z=u' subbing:
\[z'+\frac{ 4-4x }{ 1-2x}z=1-2x\]
\[e ^{\int\limits_{}^{}}\frac{ 4-4x }{ 1-2x }dx\]
The integrating factor I got is \[\frac{ e ^{2x} }{ 1-2x }\]

Answer 3

\[\frac{ e ^{2x} }{ 1-2x }(z'+\frac{ 4-4x }{ 1-2x }z)=e ^{2x}\]
\[\frac{ d }{ dx }(\frac{ e ^{2x} }{ 1-2x }z)=e ^{2x}\]
\[\frac{ e ^{2x} }{ 1-2x }z=\frac{ e ^{2x} }{ 2 }+C\]
\[z=u'=\frac{ 1-2x }{ 2 }+\frac{ C(1-2x) }{ e ^{2x} }\]
\[u=\frac{ x }{ 2 }-\frac{ x ^2 }{ 2 }+C _{2}xe ^{-2x}+C _{1}\]
But the answer in the book says
\[y=-\frac{ (2x-1)^2e^x }{ 8 }+C _{1}e^x+C _{2}xe ^{-x}\]

Answer 4

@Kainui @freckles @hartnn

Answer 5

@sammixboo @mathmale @Directrix @Zarkon

Answer 6

you know you stopped at finding u and not y right?

Answer 7

\[y=ue^{x}\]

Answer 8

\[u=\frac{x-x^2}{2}+C_2 x e^{-2x}+C_1 \\ y=ue^{x} =e^xu=e^x(\frac{x-x^2}{2}+C_2 x e^{-2x}+C_1) \\ y=\frac{x-x^2}{2}e^{x}+C_2 xe^{-2x+x}+C_1 e^{x} \\ y=\frac{x-x^2}{2}e^{x}+C_2 xe^{-2x}+C_1 e^{x}\]

Answer 9

\[-\frac{(2x-1)^2}{8}e^{x}+C_1 e^x +C_2 xe^{-x} \\ \\ -\frac{4x^2-4x+1}{8}e^{x}+C_1 e^{x}+C_2 x e^{-x} \\ -\frac{x^2-x+\frac{1}{4}}{2} e^{x}+C_1 e^{x}+C_2 x e^{-x} \\ \frac{-x^2+x-\frac{1}{4}}{2}e^{x}+C_1 e^{x}+C_2 x e^{-x} \\ \frac{x-x^2}{2} -\frac{\frac{1}{4}}{2}e^{x} +C_1 e^{x}+C_2 x e^{-x} \\ \frac{x-x^2}{2} -\frac{1}{8} e^{x} +C_1 e^{x}+C_2 x e^{-x} \\ \frac{x-x^2}{2}+(C_1 -\frac{1}{8}) e^{x}+C_2 xe^{-x} \\ \frac{x-x^2}{2}+Ke^{x}+C_2 x e^{-x}\]

so again both your y and their y are the same

Answer 10

they just through in another constant *e^x to complete the square on the polynomial part

Answer 11

Thank you!

Answer 12

threw no through :p