Author Topic: Thanksgiving bonus 4  (Read 1257 times)

Victor Ivrii

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Thanksgiving bonus 4
« on: October 05, 2018, 05:49:09 PM »
Lagrange equation is of the form
\begin{equation}
y= x\varphi (y')+\psi (y')
\label{eq1}
\end{equation}
with $\varphi(p)-p\ne 0$. To solve it we plug $p=y'$ and differentiate equation:
\begin{equation}
pdx= \varphi(p)dx + \bigl(x\varphi'(p) +\psi'(p)\bigr)dp.
\label{eq2}
\end{equation}
This is a linear ODE with respect to $x$. We find the general solution $x=f(p,C)$ and then $y=f(p,c)\varphi(p)+\psi(p)$:
\begin{equation}
\left\{\begin{aligned}
&x=f(p,C)\\
&y=f(p,c)\varphi(p)+\psi(p)
\end{aligned}\right.
\label{eq4}
\end{equation}
gives us a solution in the parametric form.

(\ref{eq1}) can have a singular solution (or solutions)
\begin{equation}
y=x\varphi(c)+\psi(c),
\label{eq5}
\end{equation}
where $c$ is a root of equation $\varphi(c)-c=0$.


Problem.
Find general and singular solutions to
$$y= 2xy'-3(y')^2.$$
« Last Edit: October 05, 2018, 06:01:49 PM by Victor Ivrii »

Monika Dydynski

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Re: Thanksgiving bonus 4
« Reply #1 on: October 06, 2018, 12:22:25 AM »
Find general and special solutions to $$y= 2xy'-3(y')^2.$$

Substituting $y'=p$, we get an equation of the form $y=2xp–3{p^2}$

Differentiating both sides, we get

$dy=2xdp+2pdx-6pdp$

$dy=pdx$   $\Rightarrow$   $pdx=2xdp+2pdx-6pdp$    $\Rightarrow$    $–pdx=2xdp–6pdp$

Dividing by $p$, we get $-dx={2x \over p}dp-6dp$   $\Rightarrow$   $ {dx \over dp} + {2 \over p}x-6=0  \tag{1}$

We get a linear differential equation. To solve $(1)$, we use an integrating factor given by $\mu=e^{\int{2 \over p}dp}=e^{2\ln|p|}=p^2$

Rewrite $(1)$ as ${2 \over p}x+{dx \over dp}=6$ and multiply by the integrating factor, $\mu(p)=p^2$ to get,

$$p^2{2 \over p}x+p^2{dx \over dp}=p^2 6$$
$${d \over dp}(p^2 x)=p^2 6$$
$$x={{2p^3+C} \over p^2}$$


Thus, the general solution to (1) is
$$x={{2p + {C \over p^2}}}\tag{2}$$

Substituting $(2)$ into the Lagrange Equation, we get


$$y={2\left({2p+\frac{C}{{{p^2}}}}\right)p–3p^2}=p^2+ \frac{2C}p$$

Thus,

\begin{equation} \left\{\begin{aligned} &x={{2p + {C \over p^2}}}\\ &y=p^2+ \frac{2C}p \end{aligned}\right. \tag{3}\end{equation}


gives us a solution in the parametric form.


The Lagrange equation, $y= 2xy'-3(y')^2$, can also have a special solution (or solutions)

$\varphi (p)-p=0$

$2p–p=0$   $\Rightarrow$    $p(2-p)=0$    $\Rightarrow$    $p_1=0$ and $p_2=2$


Thus, the special solutions are
$$y=x\varphi (0)+\psi (0)=x \cdot 0+0=0.$$
« Last Edit: October 06, 2018, 01:05:51 AM by Monika Dydynski »

Zhuojing Yu

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Re: Thanksgiving bonus 4
« Reply #2 on: October 06, 2018, 12:56:24 AM »
Here is the solution

Victor Ivrii

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Re: Thanksgiving bonus 4
« Reply #3 on: October 06, 2018, 01:31:44 AM »
Monika, you are too greedy and wasted a good problem.

Zhuojing, I give you some points despite posting after Monika, but the next time such posts will be deleted in sight: in addition of being photo rather than the scan (http://forum.math.toronto.edu/index.php?topic=1078.0) which is the first sin, they are in the wrong orientation, so I was forced to chose between rotating the monitor ot myself (kidding) which is the insult added to the injury.