Question

Reducing an differential equation to linear form

Answer 1

Cannot get this into linear form at all. I keep getting a power of 4/3

Answer 2

I've done a question like this previously, I had to use the chain rule but this equation in particular is confusing me.

Answer 3

there a nice example
http://en.wikipedia.org/wiki/Bernoulli_differential_equation

Answer 4

\[y'=2y+y^4\]
\[y'-2y=y^4;~*y^{-4}\]
\[y^{-4}y'-2y^{-3}=1\]

Answer 5

we dont want a y^-3; soo

z(y) = y(x)^-3

z(y)^-1/3 = y(x)\[-\frac{1}{3}z^{-1/4}\frac{dz}{dy}=\frac{dy}{dx}\]
start replaceing\[y^{-4}y'-2y^{-3}=1\]
\[y^{-4}y'-2z=1\]
\[y^{-4}(-\frac{1}{3}z^{-1/4}z')-2z=1\]
z^(-1/3) = y
z^(4/3) = y^-4\[-\frac{1}{3}z^{4/3}z^{-1/4}z'-2z=1\]
z^(4/3) = y^-4\[-\frac{1}{3}z^{3/3}z'-2z=1\]
z^(4/3) = y^-4\[-\frac{1}{3}z~z'-2z=1\]

Answer 6

i seem to have carried along a spurious copy paste with me lol

Answer 7

-1/3 - 3/3 = -4/3 on the derived part

Answer 8

that type of equation is called Bernoulli's DE. all type of such eqn can be changed into linear form by this technique.

Answer 9

pretty boring!!

Answer 10

\[-\frac{1}{3}z^{-4/3}z'=y'\]
\[y^{-4}y'-2z=1\]
\[-\frac{1}{3}y^{-4}z^{-4/3}z'-2z=1\]
\[-\frac{1}{3}z^{4/3}z^{-4/3}z'-2z=1\]
\[-\frac{1}{3}z'-2z=1\]thats better

typing and mathing dont mix

Answer 11

@amistre64 In your last post, how did you get -1/3 also should be u' not z'?

Answer 12

ive always did this nameing the sub as z(y) so the name really doesnt matter

Answer 13

I have y= (1/u(x))^1/3

Answer 14

u(y) = y^(-1/3) agreed?

Answer 15

since y(x) then I spose u(x) is fine, but the idea is to change it to something useable; say u(y)

Answer 16

Yes

Answer 17

we want to define y(x) not y(x)^(-3)

u(y) = y(x)^(-3) ; ^-1/3 each side

u(y)^(-1/3) = y(x) right?

Answer 18

i thought ahead and miswrote the ^-1/3 at first lol

Answer 19

ahhhh

Answer 20

\[\frac{d}{dy}(u(y)^{-1/3})=\frac{d}{dx}(y(x))\]
\[-\frac{1}{3}u(y)^{-4/3}\frac{du}{dy}=\frac{dy}{dx}\]

Answer 21

or simply\[-\frac{1}{3}u^{-4/3}~u'=y'\]

Answer 22

and now I can sub that back into the original equation

Answer 23

yes, but we still have the y^-4 to contend with\[u^{-1/3}=y;~~^ {-4}~each~side\]
\[u^{4/3}=y^{-4}\]and thats our final replacement

Answer 24

\[y^{-4}y'\to\ ~-\frac{1}{3}u^{4/3}u^{-4/3}u'\]
\[y^{-4}y'\to\ ~-\frac{1}{3}u^{4/3-4/3}u'\]
\[y^{-4}y'\to\ ~-\frac{1}{3}u'\]

Answer 25

@amistre64

Sorry, amistre, I know your solution is worked but I cannot see I wear I went wrong,

Answer 26

y^-1 is wrong, that should be 1 for starters

Answer 27

and, you need to substitute a u equivalent for y^-4

Answer 28

and replace y^-3 with u since:

u = y^-3

and

u^4/3 = (y^-3)^(4/3)
= y^-4

Answer 29

you have to replace ALL the y parts with "u" equivalents in order to reconstruct the equation in terms of "u" and not "y"