# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Thanksgiving Bonus => Topic started by: Victor Ivrii on October 05, 2018, 05:50:02 PM

Title: Thanksgiving bonus 5
Post by: Victor Ivrii on October 05, 2018, 05:50:02 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
$$2y - 4xy' - \ln (y') = 0.$$
Title: Re: Thanksgiving bonus 5
Post by: Wei Cui on October 07, 2018, 02:37:38 PM
Find general and singular solutions to:
$2y-4xy'-\ln(y')=0$

1. We transform the equation into the form:
$y=x\cdot 2y'+\frac{1}{2}\ln(y')$

We plug $p=y'$ and differentiate the equation, we get:

$pdx = 2pdx+2x\cdot dp+\frac{1}{2}\frac{1}{p}\cdot dp$

Then:
$p\frac{dx}{dp}=2p\frac{dx}{dp}+2x+\frac{1}{2p}$

$-px'-2x=\frac{1}{2p}$

$x'+\frac{2}{p}x=-\frac{1}{2p^2}$

Then $p(p) = \frac{2}{p}$, $u(p)=e^{2\int \frac{1}{p}dp} = e^{2lnp}=p^2$,
Multiply both sides with integrating factor $u(p) = p^2$ and we get:

$p^2x'+2px=-\frac{1}{2}$

$(p^2x)'=-\frac{1}{2}$

$p^2x=-\int \frac{1}{2}dp$

$p^2x=-\frac{1}{2}p+C$

$x=-\frac{1}{2p}+\frac{C}{p^2}$

Then the solution in the parametric form is: $\begin{cases}x=-\frac{1}{2p} + \frac{C}{p^2}\\ y=x\cdot 2p + \frac{1}{2}\ln p\end{cases}$

2. I am confused about the singular solution to this question. Since $\varphi(c) = 2c$ and $\varphi(c) -c=0 \implies 2c-c=0\implies c=0$.

However, for $\psi(c)$, since the domain of $\ln x$ is $x>0$. Then $\ln c=\ln 0$ will have no definition. Therefore, how am I supposed to solve the singular solution to the equation $2y-4xy'-\ln(y')=0$?
Title: Re: Thanksgiving bonus 5
Post by: Victor Ivrii on October 07, 2018, 04:57:49 PM
You do not plug $c=0$ in the general solution since the singular solution cannot be obtained from the general one by a substitution. You use (4)
Title: Re: Thanksgiving bonus 5
Post by: Wei Cui on October 07, 2018, 06:35:43 PM
Find general and singular solutions to:
$2y-4xy'-\ln(y')=0$

1. We transform the equation into the form:
$y=x\cdot 2y'+\frac{1}{2}\ln(y')$

We plug $p=y'$ and differentiate the equation, we get:

$pdx = 2pdx+2x\cdot dp+\frac{1}{2}\frac{1}{p}\cdot dp$

Then:
$p\frac{dx}{dp}=2p\frac{dx}{dp}+2x+\frac{1}{2p}$

$-px'-2x=\frac{1}{2p}$

$x'+\frac{2}{p}x=-\frac{1}{2p^2}$

Then $p(p) = \frac{2}{p}$, $u(p)=e^{2\int \frac{1}{p}dp} = e^{2lnp}=p^2$,
Multiply both sides with integrating factor $u(p) = p^2$ and we get:

$p^2x'+2px=-\frac{1}{2}$

$(p^2x)'=-\frac{1}{2}$

$p^2x=-\int \frac{1}{2}dp$

$p^2x=-\frac{1}{2}p+C$

$x=-\frac{1}{2p}+\frac{C}{p^2}$

Then the solution in the parametric form is: $\begin{cases}x=-\frac{1}{2p} + \frac{C}{p^2}\\ y=x\cdot 2p + \frac{1}{2}\ln p\end{cases}$

2. To find the singular solution, we solve the equation $\varphi(p) -p=0 \implies 2p-p=0 \implies p=0$

It follows from this that $y=C$, we can make direct substitution to make sure that the constant $C$ is equal to 0.

Therefore, the differential equation has the singular solution $y=0$.
Title: Re: Thanksgiving bonus 5
Post by: Victor Ivrii on October 07, 2018, 08:52:35 PM
No, there is no singular solution since you cannot plug $p=0$ into $\ln (p)$.