Question

(3x^2-y^2)dx+(xy-x^3y^{-1})dy=0

What technique is this??

Answer 1

http://www.symbolab.com/solver/step-by-step/%5Cleft(3x%5E%7B2%7D-y%5E%7B2%7D%5Cright)dx%2B%5Cleft(xy-x%5E%7B3%7Dy%5E%7B-1%7D%5Cright)dy%3D0/?origin=button

Answer 2

i tried separation, integrated factor, exact, bernelli

Answer 3

is this implicit diferentiation

Answer 4

i am just solving for y

Answer 5

this is in linear algebra/diff equations

Answer 6

i guess that this was implicit?

Answer 7

is this a problem in a book?

Answer 8

we have several techniques, Sepration D.E Integrated Factor, Homegeneous sub, F(ax+by) linear factor, bernalli, exact or second order D.E

Answer 9

Its for homework, our professor made it

Answer 10

siths must be typing out a real duesy

Answer 11

i hope so ^.^

Answer 12

Typically, given an ODE of this form, you want to check for exactness.
\[\begin{align*}
\underbrace{(3x^2-y^2)}_{M(x,y)}\,dx+\underbrace{\left(xy-\frac{x^3}{y}\right)}_{N(x,y)}\,dy&=0
\end{align*}\]
which has partial derivatives
\[\begin{cases}
\dfrac{\partial M}{\partial y}=-2y\\\\
\dfrac{\partial N}{\partial x}=y-\dfrac{3x^2}{y}
\end{cases}\]
Since the partial derivatives aren't equal, the equation is not exact, but it's possible that an integrating factor exists, either a function of \(x\) or \(y\) or even a function of both.

This doesn't look like a Bernoulli equation to me. Solving for the derivative gives
\[\frac{dy}{dx}=\frac{y^2-3x}{xy-\dfrac{x^3}{y}}\]
but you can't (as far as I can tell) write this in a Bernoulli form, i.e.
\[\frac{dy}{dx}+f(x)y+g(x)y^n=0\]

Answer 13

@mattyboyy

Answer 14

We might as well check to see if the function is homogeneous (I doubt it, but it doesn't hurt to check). An ODE \(\dfrac{dy}{dx}=f(x,y)\) is homogeneous if the function \(f\) satisfies
\[f(tx,ty)=t^kf(x,y)\]
for some real \(k\).
\[\begin{align*}
f(tx,ty)&=\frac{t^2y^2-3tx}{(tx)(ty)-\dfrac{t^3x^3}{ty}}\\\\
&=\frac{t^2y^2-3tx}{t^2xy-t^2\dfrac{x^3}{y}}\\\\
&=t\frac{ty^2-3x}{txy-\dfrac{x^3}{y}}\\\\
&\neq t^kf(x,y)
\end{align*}\]
and so the equation is indeed not homogeneous.

Answer 15

So do you think we need to try the not exact? put it in the form My-Mx/N to make a new equation and then find the integrating factor?

Answer 16

siths going hard with the equation button lol

Answer 17

Yeah and it is awesome!

Answer 18

We have two options for a single-variable integrating factor:
\[\mu(x)=\exp\left(\int\frac{M_y-N_x}{N}\,dx\right)\quad\text{or}\quad\mu(y)=\exp\left(\int\frac{N_x-M_y}{M}\,dy\right)\]
It looks like the first one will suffice.
\[\begin{align*}\mu(x)&=\exp\left(\int\frac{-2y-\left(y-\dfrac{3x^2}{y}\right)}{xy-\dfrac{x^3}{y}}\,dx\right)\\\\
&=\exp\left(-3\int\frac{y-\dfrac{x^2}{y}}{xy-\dfrac{x^3}{y}}\,dx\right)\\\\
&=\exp\left(-3\int\frac{y-\dfrac{x^2}{y}}{x\left(y-\dfrac{x^2}{y}\right)}\,dx\right)\\\\
&=\exp\left(-3\int\frac{dx}{x}\right)\\\\
&=\exp\left(-3\ln|x|\right)\\\\
&=\exp\left(\ln|x|^{-3}\right)\\\\
\mu(x)&=\frac{1}{x^3}
\end{align*}\]

Answer 19

wow your algebra skills are awesome

Answer 20

Thank you so much!!

Answer 21

@SithsAndGiggles College?

Answer 22

Yes. Also, I'm not using the Equation editor. All manual.

@edcion You're welcome!

Answer 23

It would have been a lot more work if I'd used the Eq button :P

Answer 24

Very interesting, you must have a lot of practice

Answer 25

after multiplying by the integrated factor the two terms are now Exact so it works thank you again!

Answer 26

You're welcome!