Question

Solve the initial-value problem y'-2y=xy^3, y(0)=2sqrt(2).

Answer 1

I did Bernoulli method and got y=+/-sqrt(4/x-16+(e^(4x))/C) and when I plug in 0 into x, I got undefined in 4/x.

Answer 2

let z = y^-2, z' = -2y^-3 y'

y^-3 y' - 2y^-2 = x

-2y^-3 y' + 4y^-2 = -2x

z' + 4z = -2x

does this look correct?

Answer 3

Yes, I got v'+4v=-2x, I did v instead of z like yours. And then the integrating factor is e^4x, right?

Answer 4

yes

Answer 5

Then I got (e^4x)v'+4ve^4x=-2xe^4x

e^(4x)v=-2*integral of xe^(4x) dx

I did integration by parts and got (xe^4x)/4-(e^4x)/16+C on the right side.

Answer 6

well, solve the homogenous

z' + 4z = 0

e^(4z)z' + 4e^(4z) z = 0

e^(4z) z = C

z = Ce^(-4z)

we solve it in z right?

Answer 7

But why did you solve for z like that?

Answer 8

y' + y = 0

e^x y implicit is x'e^x y + e^x y'

just trying to recollect is all, assuming dy/dx, then x'=1 i spose thats correct :)

Answer 9

z = Ce^(-4x)

therefore

y^-2 = Ce^(-4x)

(1/y)^2 = Ce^(-4x)

1/y = Ce^(-2x)

y = Ce^(2x)

does that look valid?

Answer 10

Let me check the answer.

Answer 11

the wolf is showing me e^(2x) is in the answer

Answer 12

The answer in the book says y=(2sqrt(2))/(sqrt(1-4x)).

Answer 13

is there an i in there someplace?

Answer 14

http://www.wolframalpha.com/input/?i=y%27-2y%3Dxy%5E3%2C++y%280%29%3D2sqrt%282%29

Answer 15

Well, the answer in the book doesn't have i, but if you plug in a number like 4 for example into x, the answer will give you a non-real result if you evaluate it in a graphing calculator.

Answer 16

trying to recall a wronskian at the moment, are at least variation of parameters

Answer 17

@myininaya

Answer 18

@satellite73 @Hero @campbell_st

Answer 19

@ganeshie8

Answer 20

\[y'-2y=xy^3\]
divide by y^3
\[y ^{-3}y \prime-2 y ^{-2}=x\]
\[put~ y ^{-2}=t,-2 y ^{-3}y \prime=\frac{ dt }{ dx }\]
\[y ^{-3}y \prime=-\frac{ 1 }{ 2 }\frac{ dt }{ dx }\]
\[-\frac{ 1 }{ 2 }\frac{ dt }{ dx }-2 t=x\]
\[\frac{ dt }{ dx }+4t=-2x\]
\[I.F.=e ^{\int\limits 4 dx}=e ^{4x}\]
c.s. is
\[t~e ^{4x}=\int\limits -2x~e ^{4x}dx+c\]
\[y ^{-2}e ^{4x}=-2~x \frac{ e ^{4x} }{ 4 }-\int\limits (-2)\frac{ e ^{4x} }{ 4 }dx+c\]
\[=-\frac{ e ^{4x} }{ 2 }+\frac{ e ^{4x} }{ 2*4 }+c\]

Answer 21

when x=0,y=2sqrt2,
find c and then substitute.