Question

Find an integrating factor for -y*dx+(x^4-x)dy=0 and solve the equation.

Answer 1

\[-y+(x^4-x)y'=0\]

Answer 2

\[\frac{ d }{ dy }(-y)=-1\]

Answer 3

usually we want a positive first term

Answer 4

thinking

we can work backwards since this is seperable to see what it would possibly be

(x^4-x)dy= y dx

dy/y= dx/(x^4-x)

ln(y) = S: [dx/(x^4-x)]

y = exp(S: [dx/(x^4-x)])

Answer 5

haha! its a variable separable!

Answer 6

But I can't solve the problem in that way. I must find an integrating factor and solve this exact equation. I already found an integrating factor for this problem, it's (x^3-1)^(-4/3).

Answer 7

y = exp(1/3 log(1-x^3)-log(x))

y = cbrt(1-x^3)/ x

Answer 8

if youve found one ... then why ask for one :)

Answer 9

Because I got stucked in the middle. Let me show all my work.

Answer 10

\[\frac{ d }{ dx }(x^4-x)=4x^3-1\]

Answer 11

for
\(y' +Py =Q\)

IF = integral e^(Px) dx

Answer 12

is it an exact equation?

Answer 13

P= 1/(x-x^4)

Answer 14

\[-y(x^3-1)^{-4/3}+(x^4-x)(x^3-1)^{-4/3}y'=0\]

Answer 15

or are we trying to find something that will make it an exact equation?

Answer 16

i didn't get how u got IF

Answer 17

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

Answer 18

\[\frac{ d }{ dx }[(x^4-x)(x^3-1)^{-4/3}]=-\frac{ 4 }{ 3 }(x^4-x)(x^3-1)^{-7/3}(3x^2)+(x^3-1)^{-4/3}(4x^3-1)\]

Answer 19

And now I'm stucked because d/dy and d/dx doesn't match, so I can't solve this.

Answer 20

I also am trying to understand how you got your integrating factor...\[\Large\rm IF=e^{\int\limits \frac{1}{x^4-x}dx}\]Which is going to be really nasty to deal with.
You already figured that part out though?

Answer 21

^^

Answer 22

Woops I guess there is a negative in there,\[\Large\rm IF=e^{\int\limits\limits \frac{1}{x-x^4}dx}\]

Answer 23

\[\int\limits_{}^{}\frac{1}{x^{4}-x} dx=\int\limits_{}^{}\frac{1}{x^4 }\frac{1}{(1-\frac{1}{x^3})}dx\]

Answer 24

now too nasty

Answer 25

Oh just a u-sub? :) haha, nice

Answer 26

awesome!

Answer 27

Then shove zedprix's negative up there after integrating

Answer 28

u = 1-1/x^3

Answer 29

\[\mu(x)=e^{\int\limits}\frac{ -1-(4x^3-1) }{ x^4-x }dx=e^{\int\limits}\frac{ -4x^3 }{ x^4-x }dx\]

Answer 30

wut? 0_o
where is that numerator coming from?

Answer 31

why are u getting 4x^3 in the numerator ?

Answer 32

\[e ^{-4\int\limits}\frac{ x^2 }{ x^3-1 }dx=(x^3-1)^{-4/3}\]

Answer 33

That's how I got the integrating factor.

Answer 34

\[-y dx+(x^4-x) dy =0 \\ \text{ Divide both sides by dx } \\ -y+(x^4-x)\frac{dy}{dx}=0 \\ \text{ Reorder terms on left hand side } \\ (x^4-x)y'-y=0 \\ \text{ divde both sides by } (x^4-x) \\ y'-\frac{1}{x^4-x}y=0 \\ y'+\frac{-1}{x^4-x}y=0 \]

Answer 35

\(y'+P(x) y =Q(x)\)
\(Integrating ~Factor = \int e^{P(x)}dx\)

Answer 36

I got 4x^3 from d/dx(x^4-x)=4x^3-1.

Answer 37

What are you trying to do it the exact way if you know what I mean?

Answer 38

Ideal has been looking at exact differential equations
I almost forgot

Answer 39

\[\mu=\exp(\int\limits_{}^{}\frac{\frac{ \partial }{ \partial y }-\frac{\partial }{ \partial x}}{x^4-x} dx) =\exp(\int\limits_{}^{} \frac{4x^3-1+1}{x^4-x} dx)\]

Answer 40

He is using harder method for this one but he is still trying to use the way that he has been learning recently.

Answer 41

http://openstudy.com/users/idealist10#/updates/543d7634e4b0b3c6e14669e4

Answer 42

u =x^3-1

Answer 43

@SithsAndGiggles

Answer 44

\[-y+(x^4-x)y'=0\]
Partials:
\[\frac{\partial }{\partial y}[-y]=-1\\
\frac{\partial }{\partial x}[x^4-x]=4x^3-1\]
IF:
\[\begin{align*}\log\mu(x)&=\int\frac{-1-(4x^3-1)}{x^4-x}~dx\\\\
&=\int\frac{4x^3}{x-x^4}~dx\\\\
&=\int\frac{4x^2}{1-x^3}~dx\\\\
&=-\frac{4}{3}\int\frac{du}{u}&\text{setting }u=1-x^3\\\\
&=-\frac{4}{3}\ln|1-x^3|\\\\
\mu(x)&=(1-x^3)^{-4/3}
\end{align*}\]
Distribute the IF:
\[-y(1-x^3)^{-4/3}-x(1-x^3)^{-1/3}y'=0\]
Partials:
\[\frac{\partial }{\partial y}[-y(1-x^3)^{-4/3}]=-(1-x^3)^{-4/3}\\
\frac{\partial }{\partial x}[-x(1-x^3)^{-1/3}]=-(1-x^3)^{-4/3}\]
Solving for \(\Psi\) might be a bit tricky...

Answer 45

Thank you so much!! @SithsAndGiggles , you made this problem so easy for me to solve!