Thread: mixed fractured equations

1. mixed fractured equations

a.
$y'+\frac{2}{x}y=\frac{y^{3}}{x^{2}}$

b.
$2y'-\frac{x}{y}=\frac{xy}{x^{2}-1}$

2. $\displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = \frac{1}{x^2}\,y^3$.

This is a Bernoulli Equation. So make the substitution $\displaystyle v = y^{-2}$.

Then $\displaystyle y = v^{-\frac{1}{2}}$ and

$\displaystyle \frac{dy}{dx} = \frac{d}{dx}(v^{-\frac{1}{2}})$

$\displaystyle = \frac{d}{dv}(v^{-\frac{1}{2}})\,\frac{dv}{dx}$

$\displaystyle = -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dx}$.

Then the DE becomes

$\displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = \frac{1}{x^2}\,y^3$

$\displaystyle -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dx} + \frac{2}{x}\,v^{-\frac{1}{2}} = \frac{1}{x^2}\,v^{-\frac{3}{2}}$

$\displaystyle \frac{dv}{dx} - \frac{4}{x}\,v = -\frac{2}{x^2}$.

This is now first order linear, so use the integrating factor method.

The integrating factor is

$\displaystyle e^{\int{-\frac{4}{x}\,dx}} = e^{-4\ln{x}} = e^{\ln{(x^{-4})}} = x^{-4}$.

Multiplying through by the integrating factor gives

$\displaystyle x^{-4}\,\frac{dv}{dx} - 4x^{-5}\,v = -2x^{-6}$

$\displaystyle \frac{d}{dx}(x^{-4}\,v) = -2x^{-6}$

$\displaystyle x^{-4}\,v = \int{-2x^{-6}\,dx}$

$\displaystyle x^{-4}\,v = \frac{2}{5}x^{-5}$

$\displaystyle v = \frac{2}{5}x^{-1}$.

Since $\displaystyle v = y^{-2}$

$\displaystyle \frac{2}{5}x^{-1} = y^{-2}$

$\displaystyle \left(\frac{2}{5}x^{-1}\right)^{-\frac{1}{2}} = y$

$\displaystyle y = \left(\frac{5}{2}\right)^{\frac{1}{2}}x^{\frac{1}{ 2}}$

$\displaystyle y = \frac{\sqrt{10x}}{2}$

3. how to solve b?

i tried to recombinate it into
(-xy^2-x^3+x)dx+2y(x^2-1)dy=0

and solve it this way

is there any other easier way
?

4. (b) is also Bernoulli. Multiply by $y$ and try letting $u = y^2.$

5. in bernulyy method we eliminate y from the right side and if the power of y is 3 then we divide by y^3
and put a new variable v=1/y^2

so we need to divide by y

but is our variable

6. Originally Posted by Danny
(b) is also Bernoulli. Multiply by $y$ and try letting $u = y^2.$
How can it be Bernoulli? When you multiply by $\displaystyle y$ you get

$\displaystyle 2y\,\frac{dy}{dx} - x = \frac{x\,y^2}{x^2 - 1}$,

which is not of the correct form $\displaystyle \frac{dy}{dx} + P(x)\,y = Q(x)\,y^n$...

7. You sure? Look closely.

8. Well I had a sign wrong, but apart from that, it does not look to be of the correct form...

9. i am confused
when i multiply by y i get a bernully for
correct?

10. Originally Posted by transgalactic
i am confused
when i multiply by y i get a bernully for
correct?
As has been said already, you multiply it by y and then make the substitution u = y^2:

$\displaystyle \frac{du}{dx} - x = \frac{x}{x^2 - 1} u$

This is a simple first order ODE solvable using the integrating factor method.

11. thanks