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Find the implicit solution to the folling DE:

$(2xy^2+3y)dx+(3x^2y+2x)dy=0$; with a starting condition $y(1)=\frac 1 2$

I can fight myself to the integrating factor.

$$y(2xy+3)dx+x(3xy+2)dy=0$$

$\mu(x,y)= \frac 1 {xy(1-xy)}$ giving us the following exact DE: $$\frac{2xy+3}{x(1-xy)}dx+\frac{3xy+2}{y(1-xy)}dy=0$$

Testing $$(\mu P)_y=\frac{2x(x(1-xy)+(2xy+3)x^2}{x^2(1-xy)^2}="some-brute-force-magicks"=\frac{1}{(1-xy)^2}$$

$$(\mu Q)_x=\frac{3y(1-xy)y-(3xy+2)y(-y)}{y^2(1-xy)^2}=\frac{1}{(1-xy)^2}$$

Thus $(\mu P)_y=(\mu Q)_x$

From here I try to get $u=?$ $$u=\int(\mu P)dx=\int\frac{2xy+3}{x(1-xy)}dx=3ln(x)-5ln(1-xy)+C(y)$$

So far so good but ...now ...either this so far is wrong in theory, I messed up my integration, or I keep misscalculating my deritative as my calculation $u_y \neq (\mu Q)$

I tried (whats the English word) rearranging (?) $$u=ln \left(\frac{x^3}{(1-xy)^5} \right)+C$$

Help would be nice.

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    $\begingroup$ $u$ is correct, but write $C(y)$ as $N(y)$. Now, take the derivative of $u_y$ and substitute $\dfrac{3xy+2}{y(1-xy)}$ and solve for $N$. $\endgroup$ – Moo Feb 12 '17 at 2:34
  • $\begingroup$ @Moo The thing is if I compare my $u_y$ to $(\mu Q)$ I get such a disgusting value for $C'(y)$ So im not sure if I should have put C under the log via $u=log(...)+logC(y)$ or not. On top of that I'm not fully sure how to even get to the $y(x)$ from here on to even enter the boundry condition of $y(1)=1/2$ in $\endgroup$ – Katpton Liamfuppinshire Feb 12 '17 at 3:21
  • $\begingroup$ I added an answer wit more details. $\endgroup$ – Moo Feb 12 '17 at 3:22
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Solve

$$\tag 1 M(x, y)~ dx + N(x, y)~ dy = (2xy^2+3y)~dx+(3x^2y+2x)~dy=0,y(1)=\dfrac 1 2$$

You found an integrating factor that makes $M_y = N_x$ as

$$\tag 2 \mu(x,y)= \dfrac 1 {xy(1-xy)}$$

Multiplying $(1)$ by $(2)$ gives the exact DEQ

$$\tag 3 \dfrac{2xy+3}{x(1-xy)}dx+\dfrac{3xy+2}{y(1-xy)}dy=0$$

Now

$$g(x, y) = \displaystyle \int \left(\dfrac{2xy+3}{x(1-xy)}\right)~dx = 3 \ln x - 5 \ln(1 - x y) + h(y)$$

We can now write

$$\dfrac{\partial g(x, y)}{\partial y} = \dfrac{5 x}{1-x y } + h'(y) = \dfrac{3xy+2}{y(1-xy)} \implies h'(y) = \dfrac{3xy+2}{y(1-xy)} - \dfrac{5 x}{1-x y } = \dfrac{2}{y}$$

Solving

$$h(y) = 2 \ln y$$

We now have

$$g(x, y) = 3 \ln x - 5 \ln(1 - x y) + h(y) = 3 \ln x - 5 \ln(1 - x y) + 2 \ln y = c$$

I will assume you can take it from here to solve for $c$ using the IC and get the final implicit form.

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  • $\begingroup$ Hmmm ya i noticed where i did my mistake. And it was always the same when i tried calculating $g_y$ So from here I can simply enter $x=1$ and $y=\frac1 2$ ? To get my C out of this? Might stuggle a bit with the implicit form here. Was never good at this part. EDIT: And thanks, now I can finally try and get some sleep. $\endgroup$ – Katpton Liamfuppinshire Feb 12 '17 at 3:27
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    $\begingroup$ Yes, just sub in those values of $x$ and $y$, get $c$ and you are done. The solution is already in implicit form - just find $c$ and replace in the implicit form solution. $\endgroup$ – Moo Feb 12 '17 at 3:29

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