First-order nonlinear ODE with a parameter

**Ment** · January 14th 2010, 06:22 AM

Help me with this equation with a parameter $a$ . I can not even guess how I begin to solve it.

$y^3dy - (y^2+ax)dx = 0$

Thanks!

**DeMath** · January 14th 2010, 10:02 AM

Originally Posted by Ment

Help me with this equation with a parameter $a$ . I can not even guess how I begin to solve it.

$y^3dy - (y^2+ax)dx = 0$

Thanks!

$y^3\,dy - \left(y^2+a x\right)\,dx = 0 ~ \Leftrightarrow ~ y^3y' - y^2 - a x = 0 ~ \Leftrightarrow$

$\Leftrightarrow ~ y^2\left(y^2\right)^\prime - 2y^2 - 2a x = 0 ~ \Leftrightarrow ~ \left\{\begin{gathered}y^2 = xp, \hfill \\ \left(y^2\right)^\prime = p + xp' \hfill \\ \end{gathered}\right\} ~ \Leftrightarrow$

$\Leftrightarrow ~ xp\left(p + xp'\right) - 2xp - 2ax = 0 ~ \Leftrightarrow ~ xpp' = 2a + 2p - p^2 ~ \Leftrightarrow$

$\Leftrightarrow ~ \frac{dx}{x} = \frac{p\,dp} {2a + 2p - p^2} ~ \Leftrightarrow ~ \ln|x| + C = \int \frac{p\,dp} {2a + 2p - p^2}.$

Now consider three cases:

#1. $a \in \left(-\infty;\,- \frac{1}{2}\right)$

$\int {\frac{{p\,dp}}{{ - 2a + 2p - {p^2}}}} = - \frac{1}{2}\int {\frac{{2 - 2p - 2}}{{ - 2a + 2p - {p^2}}}\,dp} =$

$= - \frac{1} {2}\int {\frac{{d\left( { - 2a + 2p - {p^2}} \right)}} {{ - 2a + 2p - {p^2}}}} - \int {\frac{{dp}} {{2a - 1 + {{\left( {p - 1} \right)}^2}}}} =$

$= - \frac{1}{2}\ln\left|p^2 - 2p + 2a\right| - \frac{1}{{\sqrt {2a - 1} }}\int \frac{d\left(\dfrac{p - 1}{\sqrt{2a - 1}}\right)}{{1 + {{\left( {\dfrac{{p - 1}}{{\sqrt {2a - 1} }}} \right)}^2}}} =$

$= -\frac{1}{2}\ln \left|p^2 - 2p + 2a\right| - \frac{1}{\sqrt{2a-1}}\arctan\frac{p-1}{\sqrt{2a-1}} + C.$

So, when $a \in \left(-\infty;\,- \frac{1}{2}\right)$ , you have this solution

$\ln|x| + C = - \frac{1}{2}\ln\left|\frac{y^4}{x^2} - \frac{2y^2}{x} + 2a\right| - \frac{1}{\sqrt{2a - 1}}\arctan \frac{\dfrac{y^2}{x} - 1}{\sqrt{2a - 1}} ~ \Leftrightarrow$

$\Leftrightarrow ~ \frac{1}{2}\ln\left|y^4 - 2xy^2 + 2ax^2\right| + \frac{1}{\sqrt{2a - 1}}\arctan \frac{y^2 - x}{x\sqrt{2a - 1}} = C.$

#2. If $a =-\frac{1}{2}$ then

$\int \frac{pdp}{-1 + 2p - p^2} = - \int \frac{p\,dp}{(p - 1)^2} = - \int \frac{1+p-1}{(p-1)^2}\,dp =$

$= - \int \frac{dp}{(p - 1)^2} - \int \frac{dp}{p - 1} = \frac{1} {p - 1} - \ln|p - 1| + C.$

So, when $a =-\frac{1}{2}$ , you have this solution

$\ln|x| + C = \frac{1}{\dfrac{y^2}{x} - 1} - \ln\left|\frac{y^2}{x} - 1\right| ~ \Leftrightarrow ~ \frac{x}{y^2 - x} - \ln \left|y^2 - x\right| = C.$

#3. If $a \in \left(-\frac{1}{2};\,+\infty\right)$ then

$\int\frac{p\,dp}{2a + 2p - p^2} = - \frac{1}{2}\int \frac{2 - 2p - 2}{2a + 2p - p^2}\,dp =$

$= \int\frac{dp}{2a + 2p - p^2} - \frac{1}{2}\int \frac{d\left(2a + 2p - p^2\right)}{2a + 2p - p^2} =$

$= \int\frac{dp}{2a + 1 - \left( p - 1\right)^2} - \frac{1}{2}\ln\left|2a + 2p - p^2\right| =$

$= \frac{1}{\sqrt{2a + 1}}\int \frac{d\left(\dfrac{p - 1}{\sqrt{2a + 1}}\right)}{1 - \left(\dfrac{p - 1}{\sqrt {2a + 1}}\right)^2} - \frac{1}{2}\ln \left|2a + 2p - p^2\right| =$

$= \frac{1}{2\sqrt{2a + 1}}\ln\left|\frac{p - 1 + \sqrt{2a + 1}}{p - 1 - \sqrt {2a + 1}}\right| - \frac{1}{2}\ln \left|2a + 2p - p^2\right| + C.$

So, when $a \in \left(-\frac{1}{2};\,+\infty\right)$ , you have this solution

$\ln \left| x \right| + C = \frac{1} {{2\sqrt {2a + 1} }}\ln \left| {\frac{{\dfrac{{{y^2}}} {x} - 1 + \sqrt {2a + 1} }} {{\dfrac{{{y^2}}} {x} - 1 - \sqrt {2a + 1} }}} \right| - \frac{1} {2}\ln \left| {2a + \frac{{2{y^2}}} {x} - \frac{{{y^4}}} {{{x^2}}}} \right| ~ \Leftrightarrow$

$\Leftrightarrow ~ \frac{1} {{2\sqrt {2a + 1} }}\ln \left| {\frac{{{y^2} - \left( {1 - \sqrt {2a + 1} } \right)x}} {{{y^2} - \left( {1 + \sqrt {2a + 1} } \right)x}}} \right| - \frac{1} {2}\ln \left| {2a{x^2} + 2x{y^2} - {y^4}} \right| = C.$

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