Question

Solve the initial value problem (x^2-4)y"+4xy'+2y=x+2, y(0)=-1/3, y'(0)=-1, given that y1=1/(x-2) satisfies the complementary equation.

Answer 1

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

Answer 2

Simplify,
\[u''+\frac{ 2 }{ x+2 }u'=1\]
z=u' subbing:
\[z'+\frac{ 2 }{ x+2 }z=1\]
\[(x+2)^2z=\frac{ (x+3)^3 }{ 3 }+C\]
\[z=u'=\frac{ x+2 }{ 3 }+C(x+2) ^{-2}\]
integrate
\[u=\frac{ x^2 }{ 6 }+\frac{ 2x }{ 3 }-C _{1}(x+2) ^{-1}+C _{2}\]
Since \[y=\frac{ u }{ x-2 }\],
\[y=\frac{ x^2+4x }{ 6(x-2) }-\frac{ C _{1} }{ x^2-4 }+\frac{ C _{2} }{ x-2 }\]
plug in y(0)=-1/3,
-1/3=C1/4-C2/2
At the end, I didn't get the right answer. The answer in the book says \[y=\frac{ (x+2)^2 }{ 6(x-2) }+\frac{ 2x }{ x^2-4 }\]

Answer 3

@Kainui @Hero

Answer 4

Are you sure they're really different and not just different ways of arranging them by algebra?

Answer 5

Well, the constants I got are C1=4 and C2=8/3. I don't know if those are right and if the book's answer is right.

Answer 6

@freckles

Answer 7

i have to go with @kainui on this one... have you checked to see if both answers are the same...
try to write their answer as your answer
\[y=\frac{(x+2)^2}{6(x-2)}+\frac{2x}{x^2-4} \\ y=\frac{x^2+4x+4}{6(x-2)}+\frac{2x}{x^2-4} \\ y=\frac{x^2+4x}{6(x-2)}+ \frac{4}{6(x-2)}+\frac{2x}{x^2-4} \\ y=\frac{x^2+4x}{6(x-2)}+\frac{2}{3(x-2)}+\frac{1}{x-2}+\frac{1}{x+2} \\ y=\frac{x^2+4x}{6(x-2)}+\frac{2}{3(x-2)}+\frac{6}{3(x-2)} -\frac{1}{x-2}+\frac{1}{x+2} \\ y=\frac{x^2+4x}{6(x-2)} +\frac{8}{3(x-2)}+\frac{1}{x+2}-\frac{1}{x-2}\]
...
see if you can continue
there is only one more step really

Answer 8

though you could have integrated differently to get their answer without undoing and doing stuff to see if their answer is your answer

Answer 9

\[u'=\frac{x+2}{3}+C(x+2)^{-2} \\ u=\frac{(x+2)^2}{3 \cdot 2}-C(x+2)^{-1} +D \\ u=\frac{(x+2)^2}{6}-C(x+2)^{-1}+D \\ \text{ instead of writing } \\ u'=\frac{x}{3}+\frac{2}{3}+C(x+2)^{-2} \\ \text{ and the integrating which gives } \\ u=\frac{x^2}{6}+\frac{2}{3}x-C(x+2)^{-1}+D\]
neither way is wrong

Answer 10

But are the constants I got right or wrong? I got C2=8/3 and C1=4, when I plug them into y, I got a weird answer.

Answer 11

you don't know where to continue from the book's answer above?

Answer 12

\[y=\frac{(x+2)^2}{6(x-2)}+\frac{2x}{x^2-4} \\ y=\frac{x^2+4x+4}{6(x-2)}+\frac{2x}{x^2-4} \\ y=\frac{x^2+4x}{6(x-2)}+ \frac{4}{6(x-2)}+\frac{2x}{x^2-4} \\ y=\frac{x^2+4x}{6(x-2)}+\frac{2}{3(x-2)}+\frac{1}{x-2}+\frac{1}{x+2} \\ y=\frac{x^2+4x}{6(x-2)}+\frac{2}{3(x-2)}+\frac{6}{3(x-2)} -\frac{1}{x-2}+\frac{1}{x+2} \\ y=\frac{x^2+4x}{6(x-2)} +\frac{8}{3(x-2)}+\frac{1}{x+2}-\frac{1}{x-2} \]
hint try combing the last two fractions

Answer 13

It's 2x/(x^2-4).

Answer 14

combining the last two fractions?

Answer 15

\[\frac{1}{x+2}-\frac{1}{x-2}=\frac{(x-2)-(x+2)}{(x+2)(x-2)}=\frac{x-x-2-2}{x^2-4}=\frac{-4}{x^2-4}\]

Answer 16

you should see the book's answer is not equivalent to your answer

Answer 17

\[y=\frac{x^2+4x}{6(x-2)}+\frac{8}{3(x-2)}-\frac{4}{x^2-4} \\ =\frac{(x+2)^2}{6(x-2)}+\frac{2x}{x^2-4}\]

Answer 18

as shown above

Answer 19

You know what? I got the right answer. I found my mistake.
\[u'=\frac{ x+2 }{ 3 }+C(x+2) ^{-2}\]
integrate, it should be
\[u=\frac{ (x+2)^2 }{ 6 }-C _{1}(x+2) ^{-1}+C _{2}\]
, not
\[u=\frac{ x^2 }{ 6 }+\frac{ 2x }{ 3 }-C _{1}(x+2) ^{-1}+C _{2}\]

Answer 20

you didn't make a mistake

Answer 21

I showed their answer is same as yours above :p

Answer 22

also you can write (x+2)/3 as x/3+2/3
and integrate if you want
or straight up integrate (x+2)/3

Answer 23

I just want to say, thank you! :)

Answer 24

np

Answer 25

i see why there was confusion now
"you should see the book's answer is not equivalent to your answer
"
this sentence was suppose to say now equivalent instead of not equivalent
but anyways the work alone should have reflected the now instead of the not :p