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 ibanez1608 November 27th, 2016 09:19 AM

help separate eq

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 topsquark November 27th, 2016 10:47 AM

Quote:
 Originally Posted by ibanez1608 (Post 554793) please help i dont know how to start separating the eq
$\displaystyle xyy' = (x + y)^2$ for y(1) = 0.

You might be able to separate this (you'll need a change of variables if you can even do it) but note that your initial condition isn't part of the solution set:
$\displaystyle xyy' = 1 \cdot 0 \cdot y'(1) = 0$ for any value of y'(1)

but
$\displaystyle (x + y)^2 = (1 + 0)^2 = 1$

-Dan

 Prove It November 27th, 2016 03:18 PM

\displaystyle \begin{align*} x\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} &= \left( x + y \right) ^2 \\ &= x^2 + 2\,x\,y + y^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{x}{y} + 2 + \frac{y}{x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\frac{y}{x}} + 2 + \frac{y}{x} \end{align*}

Now let \displaystyle \begin{align*} u = \frac{y}{x} \implies y = u\,x \end{align*}, then \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = u + x\,\frac{\mathrm{d}u}{\mathrm{d}x} \end{align*} giving

\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\frac{y}{x}} + 2 + \frac{y}{x} \\ u + x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{u} + 2 + u \\ x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{u} + 2 \\ x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1 + 2\,u}{u} \\ \frac{u}{1 + 2\,u}\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{x} \\ \int{ \frac{u}{1 + 2\,u}\,\frac{\mathrm{d}u}{\mathrm{d}x} \,\mathrm{d}x} &= \int{\frac{1}{x}\,\mathrm{d}x} \\ \int{ \frac{u}{1 + 2\,u} \,\mathrm{d}u} &= \ln{ \left| x \right| } + C_1 \\ \frac{1}{2} \int{ \frac{2\,u}{1 + 2\,u} \,\mathrm{d}u } &= \ln{ \left| x \right| } + C_1 \\ \frac{1}{2} \int{ \left( 1 - \frac{1}{1 + 2\,u} \right) \,\mathrm{d}u } &= \ln{ \left| x \right| } +C_1 \\ \frac{1}{4} \int{ \left( 2 - \frac{2}{1 + 2\,u} \right) \,\mathrm{d}u } &= \ln{ \left| x \right| } + C_1 \\ \frac{1}{4} \, \left( 2\,u - \ln{ \left| 1 + 2\,u \right| } \right) + C_2 &= \ln{ \left| x \right| } + C_1 \\ \frac{1}{2}\,u - \frac{1}{4}\,\ln{ \left| 1 + 2\,u \right| } &= \ln{ \left| x \right| } + C \textrm{ where } C = C_1 - C_2 \\ \frac{y}{2\,x} - \frac{1}{4}\, \ln{ \left| 1 + \frac{2\,y}{x} \right| } &= \ln{ \left| x \right| } + C \end{align*}

Now since \displaystyle \begin{align*} y \left( 1 \right) = 0 \end{align*} that means

\displaystyle \begin{align*} \frac{0}{2\cdot 1} - \frac{1}{4} \,\ln{ \left| 1 + \frac{2\cdot 0}{1} \right| } &= \ln{ \left| 1 \right| } + C \\ 0 - \frac{1}{4}\, \ln{\left( 1 \right) } &= \ln{ \left( 1 \right) } + C \\ C &= 0 \end{align*}

and thus \displaystyle \begin{align*} \frac{y}{2\,x} - \frac{1}{4}\, \ln{ \left| 1 + \frac{2\,y}{x} \right| } &= \ln{ \left| x \right| } \end{align*}.

 topsquark November 27th, 2016 03:31 PM

Quote:
 Originally Posted by Prove It (Post 554862) Now since \displaystyle \begin{align*} y \left( 1 \right) = 0 \end{align*} that means \displaystyle \begin{align*} \frac{0}{2\cdot 1} - \frac{1}{4} \,\ln{ \left| 1 + \frac{2\cdot 0}{1} \right| } &= \ln{ \left| 1 \right| } + C \\ 0 - \frac{1}{4}\, \ln{\left( 1 \right) } &= \ln{ \left( 1 \right) } + C \\ C &= 0 \end{align*} and thus \displaystyle \begin{align*} \frac{y}{2\,x} - \frac{1}{4}\, \ln{ \left| 1 + \frac{2\,y}{x} \right| } &= \ln{ \left| x \right| } \end{align*}.
Nicely done. But I still object to the initial condition. In order to find the solution to the DEq you divided both sides by xy. So y(1) = 0 is still outside the solution set.

-Dan

 skipjack November 27th, 2016 08:00 PM

The graph for the equation obtained contains the point (1, 0), but doesn't have a slope there, so the differential equation isn't satisfied there. Hence there is no solution.

 ibanez1608 November 28th, 2016 04:35 AM

ty prove it

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