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Solving $\frac{d^2y}{dx^2} = \frac{y}{x}\cdot\frac{dy}{dx}.$

Even though ‘contrib_ode’, Maxima’s ODE solver choked on this equation (see “An Alternate Solver of ODEs“), it still can be solved as demonstrated below:

multiplied by $x,$

$x\frac{d^2y}{dx^2} = y\frac{dy}{dx},$

i.e.,

$x\frac{d}{dx}\left(\frac{dy}{dx}\right) = y\frac{dy}{dx}.$

Integrate it, we have

$\int x\frac{d}{dx}\left(\frac{dy}{dx}\right)\;dx = \int y \frac{dy}{dx} \;dx \implies \int \left(\frac{dy}{dx}\right)'\cdot x\;dx = \frac{1}{2}y^2+c.\quad\quad\quad(1-1)$

By $\int u'\cdot v\;dx = u\cdot v - \int u\cdot v'\;dx$ (see “Integration by Parts Done Right“),

$\int \left(\frac{dy}{dx}\right)'\cdot x\;dx = \frac{dy}{dx}\cdot x-\int\frac{dy}{dx}\cdot x'\;dx \overset{x'=1}{=} x\frac{dy}{dx}-\int \frac{dy}{dx}\;dx = x\frac{dy}{dx}-y.$

As a result, (1-1) yields a new ODE

$x\frac{dy}{dx} -y = \frac{1}{2}y^2 + c\quad\quad\quad(1-2)$

or,

$2x\frac{dy}{dx}=2y+y^2+c_1,\quad c_1=2c.\quad\quad\quad(1-3)$

Upon submitting (1-3) to Omega CAS Explorer in non-expert mode, the CAS asks for the range of $c_1.$

Fig. 1

Depending on the range provided, ‘ode2’ gives three different solutions (see Fig. 2, 3 and 4).

Fig. 2 $c_1>1$

Fig. 3 $c_1<1$

Fig. 4 $c_1=1$

Let’s also solve (1-3) manually:

If $y^2+2y+c_1=0,$ (1-3) has a constant solution

$y=c_2.\quad\quad\quad(2-1)$

In fact, this solution can be observed from $\frac{d^2y}{dx^2} = \frac{y}{x}\cdot\frac{dy}{dx}$ right away.

Otherwise ( $y^2+2y+c_1 \ne 0$ ),

$\frac{1}{y^2+2y+c_1}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

That is,

$\frac{1}{y^2+2y+1+c_1-1}\cdot\frac{dy}{dx}=\frac{1}{2x}$

$\frac{1}{(y+1)^2+c_1-1}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

For $c_1-1>0,$ let $k=\sqrt{c_1-1},$ we have

$\frac{1}{(y+1)^2+k^2}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

Divide both numerator and denominator on the left side by $k^2$ ,

$\frac{1}{k^2}\cdot\frac{1}{(\frac{y+1}{k})^2+1}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

Write it as

$\frac{1}{k}\cdot\frac{1}{k}\cdot\frac{1}{(\frac{y+1}{k})^2+1}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

Multiply both sides by $k,$

$\frac{1}{k}\cdot\frac{1}{(\frac{y+1}{k})^2+1}\cdot\frac{dy}{dx}=k\cdot\frac{1}{2x}$

Integrate it,

$\int\frac{1}{k}\cdot\frac{1}{(\frac{y+1}{k})^2+1}\cdot\frac{dy}{dx}\;dx=\int k\cdot\frac{1}{2x}\;dx,$

we obtain

$\arctan(\frac{y+1}{k}) = \frac{k}{2}\log(|x|)+c_2.$

i.e.,

$\arctan(\frac{y+1}{\sqrt{c_1-1}}) = \frac{\sqrt{c_1-1}}{2}\log(|x|)+c_2$

or,

$y = 2k_1\tan(k_1\log|x| +k_2) -1\quad\quad\quad(2-2)$

where $k_1=\frac{\sqrt{c_1-1}}{2}, k_2=c_2.$

For $c_1-1<0,$ let $k=\sqrt{1-c_1},$

$\frac{1}{(y+1)^2-k^2}\cdot\frac{dy}{dx}=\frac{1}{2x}$

$\frac{1}{(y+1-k)\cdot(y+1+k)}\cdot\frac{dy}{dx}=\frac{1}{2x}$

$\frac{1}{2k}\left(\frac{1}{y+1-k} - \frac{1}{y+1+k}\right)\cdot\frac{dy}{dx}= \frac{1}{2x}$

$\int \left(\frac{1}{y+1-k} - \frac{1}{y+1+k}\right)\cdot\frac{dy}{dx}\;dx= \int\frac{k}{x}\;dx$

$\log|y+1-k| - \log|y+1+k| = k\cdot\log|x| + c_2$

$\log\bigg|\frac{y+1-k}{y+1+k}\bigg| = k\cdot\log|x| + c_2$

$\log\bigg|\frac{y+1-\sqrt{1-c_1}}{y+1+\sqrt{1-c_1}}\bigg| = \sqrt{1-c_1}\cdot\log|x| + c_2$

$y= -\frac{k_2(1+k_1)|x|^{k_1}+k_1-1}{k_2|x|^{k_1}-1}, \quad k_1=\sqrt{1-c_1}, k_2=e^{c_2}.\quad\quad\quad(2-3)$

For $c_1-1=0 \quad(c_1=1),$

$\frac{1}{(y+1)^2}\cdot\frac{dy}{dx}=\frac{1}{2x}$

$\int \frac{1}{(y+1)^2}\cdot\frac{dy}{dx}\;dx=\int \frac{1}{2x}\;dx$

$-\frac{1}{y+1} = \frac{1}{2}\log|x| +c_2$

$y+1 = \frac{-1}{\frac{1}{2}\log|x|+c_2}$

$y = -1 - \frac{1}{\frac{1}{2}\log|x|+c_2}.\quad\quad\quad(2-4)$

Notice when $c_1 = 0, c=0.$ (1-2) becomes

$\frac{dy}{dx} - \frac{1}{x}y = \frac{1}{2x}y^2.$

This is a Bernoulli’s Equation $\frac{dy}{dx}+f(x)y=g(x)y^{\alpha}$ with $f(x)=\frac{1}{x}, g(x)=\frac{1}{2x}$ and $\alpha=2.$ Solving it (see “Meeting Mr. Bernoulli“),