| Version 9 |
Version 8 |
| There are certain \PMlinkescapetext{types} of non-linear ordinary differential equations of \PMlinkname{first order}{ODE} which may by a suitable substitution be \PMlinkescapetext{reduced} to a form where one can \PMlinkname{separate}{SeparationOfVariables} the variables.\\ |
There are certain \PMlinkescapetext{types} of non-linear ordinary differential equations of \PMlinkname{first order}{ODE} which may by a suitable substitution be \PMlinkescapetext{reduced} to a form where one can \PMlinkname{separate}{SeparationOfVariables} the variables.\\ |
|
|
| \textbf{I.\; So-called homogeneous differential equation} |
\textbf{I.\; So-called homogeneous differential equation} |
|
|
| This means the equation of the form |
This means the equation of the form |
| $$X(x,\,y)dx+Y(x,\,y)dy = 0,$$ |
$$X(x,\,y)dx+Y(x,\,y)dy = 0,$$ |
| where $X$ and $Y$ are two homogeneous functions of the same \PMlinkname{degree}{HomogeneousFunction}.\, Therefore, if the equation is written as |
where $X$ and $Y$ are two homogeneous functions of the same \PMlinkname{degree}{HomogeneousFunction}.\, Therefore, if the equation is written as |
| $$\frac{dy}{dx} = -\frac{X(x,\,y)}{Y(x,\,y)},$$ |
$$\frac{dy}{dx} = -\frac{X(x,\,y)}{Y(x,\,y)},$$ |
| its right hand side is a homogeneous function of degree 0, i.e. it depends only on the ratio $y\!:\!x$, and has thus the form |
its right hand side is a homogeneous function of degree 0, i.e. it depends only on the ratio $y\!:\!x$, and has thus the form |
| \begin{align} |
\begin{align} |
| \frac{dy}{dx} = f\left(\frac{y}{x}\right). |
\frac{dy}{dx} = f\left(\frac{y}{x}\right). |
| \end{align} |
\end{align} |
|
|
| Accordingly, if this ratio is constant, then also $\frac{dy}{dx}$ is constant; thus all lines \, $\frac{y}{x} =$ constant\, are isoclines of the family of the integral curves which intersect any such line isogonally. |
Accordingly, if this ratio is constant, then also $\frac{dy}{dx}$ is constant; thus all lines \, $\frac{y}{x} =$ constant\, are isoclines of the family of the integral curves which intersect any such line isogonally. |
|
|
| We can infer as well, that if one integral curve is represented by\, $x = x(t)$,\; $y = y(t)$,\, then also\, $x = Cx(t)$,\; $y = Cy(t)$\, \PMlinkescapetext{represents} an integral curve for any constant $C$.\, Hence the integral curves are homothetic with respect to the origin; therefore some people call the equation (1) a {\em similarity equation}. |
We can infer as well, that if one integral curve is represented by\, $x = x(t)$,\; $y = y(t)$,\, then also\, $x = Cx(t)$,\; $y = Cy(t)$\, \PMlinkescapetext{represents} an integral curve for any constant $C$.\, Hence the integral curves are homothetic with respect to the origin; therefore some people call the equation (1) a {\em similarity equation}. |
|
|
| For generally solving the equation (1), make the substitution |
For generally solving the equation (1), make the substitution |
| $$\frac{y}{x} := t; \quad y = tx; \quad \frac{dy}{dx} = t+x\frac{dt}{dx}.$$ |
$$\frac{y}{x} := t; \quad y = tx; \quad \frac{dy}{dx} = t+x\frac{dt}{dx}.$$ |
| The equation takes the form |
The equation takes the form |
| \begin{align} |
\begin{align} |
| t+x\frac{dt}{dx} = f(t) |
t+x\frac{dt}{dx} = f(t) |
| \end{align} |
\end{align} |
| which shows that any \PMlinkname{root}{Equation} $t_\nu$ of the equality \,$f(t) = t$\, gives a singular solution \, $y = t_\nu x$. |
which shows that any \PMlinkname{root}{Equation} $t_\nu$ of the equality \,$f(t) = t$\, gives a singular solution \, $y = t_\nu x$. |
| The variables in (2) may be \PMlinkescapetext{separated}: |
The variables in (2) may be \PMlinkescapetext{separated}: |
| $$\frac{dx}{x} = \frac{dt}{f(t)\!-\!t}$$ |
$$\frac{dx}{x} = \frac{dt}{f(t)\!-\!t}$$ |
| Thus one obtains\, $\ln{|x|} = \int\!\frac{dt}{f(t)\!-\!t}+ \ln{C}$, whence the general solution of the homogeneous differential equation (1) is in a parametric form |
Thus one obtains\, $\ln{|x|} = \int\!\frac{dt}{f(t)\!-\!t}+ \ln{C}$, whence the general solution of the homogeneous differential equation (1) is in a parametric form |
| $$x = Ce^{\int\!\frac{dt}{f(t)\!-\!t}}, \quad y = Cte^{\int\!\frac{dt}{f(t)\!-\!t}}.$$ |
$$x = Ce^{\int\!\frac{dt}{f(t)\!-\!t}}, \quad y = Cte^{\int\!\frac{dt}{f(t)\!-\!t}}.$$ |
|
|
|
|
|
|
| \textbf{II.\; Equation of the form\, {\em y}$'${\em = f(ax+by+c)}} |
\textbf{II.\; Equation of the form\, {\em y}$'${\em = f(ax+by+c)}} |
|
|
| It's a question of the equation |
It's a question of the equation |
| \begin{align} |
\begin{align} |
| \frac{dy}{dx} = f(ax+by+c), |
\frac{dy}{dx} = f(ax+by+c), |
| \end{align} |
\end{align} |
| where $a$, $b$ and $c$ are given constants.\, If\, $ax+by$ is constant, then $\frac{dy}{dx}$ is constant, and we see that the lines\, $ax+by =$ constant\, are isoclines of the intgral curves of (3). |
where $a$, $b$ and $c$ are given constants.\, If\, $ax+by$ is constant, then $\frac{dy}{dx}$ is constant, and we see that the lines\, $ax+by =$ constant\, are isoclines of the intgral curves of (3). |
|
|
| Let |
Let |
| \begin{align} |
\begin{align} |
| ax+by+c := u |
ax+by+c := u |
| \end{align} |
\end{align} |
| be a new variable.\, It changes the equation (3) to |
be a new variable.\, It changes the equation (3) to |
| \begin{align} |
\begin{align} |
| \frac{du}{dx} = a+bf(u). |
\frac{du}{dx} = a+bf(u). |
| \end{align} |
\end{align} |
| Here, one can see that the real zeros $u$ of the right hand side yield lines (4) which are integral curves of (3), and thus we have singular solutions.\, Moreover, one can separate the variables in (5) and integrate, obtaining $x$ as a function of $u$.\, Using still (4) gives also $y$.\, The general solution is |
Here, one can see that the real zeros $u$ of the right hand side yield lines (4) which are integral curves of (3), and thus we have singular solutions.\, Moreover, one can separate the variables in (5) and integrate, obtaining $x$ as a function of $u$.\, Using still (4) gives also $y$.\, The general solution is |
| $$x = \int\frac{du}{a+bf(u)}+C, \quad y = \frac{1}{b}\left(u-c-a\int\frac{du}{a+bf(u)}-aC\right).\\$$ |
$$x = \int\frac{du}{a+bf(u)}+C, \quad y = \frac{1}{b}\left(u-c-a\int\frac{du}{a+bf(u)}-aC\right).\\$$ |
|
|
|
|
| \textbf{Example.}\, In the nonlinear equation |
\textbf{Example.}\, In the nonlinear equation |
| $$\frac{dy}{dx} = (x-y)^2,$$ |
$$\frac{dy}{dx} = (x-y)^2,$$ |
| which is of the type II, one cannot separate the variables $x$ and $y$.\, The substitution\, $x-y := u$\, converts it to |
which is of the type II, one cannot separate the variables $x$ and $y$.\, The substitution\, $x-y := u$\, converts it to |
| $$\frac{du}{dx} = 1-u^2,$$ |
$$\frac{du}{dx} = 1-u^2,$$ |
| where one can separate the variables.\, Since the right hand side has the zeros\, $u = \pm1$,\, the given equation has the singular solutions $y$ given by\, $x-y = \pm1$.\, Separating the variables $x$ and $u$, one obtains |
where one can separate the variables.\, Since the right hand side has the zeros\, $u = \pm1$,\, the given equation has the singular solutions $y$ given by\, $x-y = \pm1$.\, Separating the variables $x$ and $u$, one obtains |
| $$dx = \frac{du}{1-u^2},$$ |
$$dx = \frac{du}{1-u^2},$$ |
| whence |
whence |
| $$x = \int\frac{du}{(1+u)(1-u)} = \frac{1}{2}\int\left(\frac{1}{1+u}+\frac{1}{1-u}\right)du |
$$x = \int\frac{du}{(1+u)(1-u)} = \frac{1}{2}\int\left(\frac{1}{1+u}+\frac{1}{1-u}\right)du |
| = \frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C.$$ |
= \frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C.$$ |
| Accordingly, the given differential equation has the parametric solution |
Accordingly, the given differential equation has the parametric solution |
| $$x = \ln\sqrt{\left|\frac{1\!+\!u}{1\!-\!u}\right|}+C, \quad y = \ln\sqrt{\left|\frac{1\!+\!u}{1\!-\!u}\right|}-u\!+\!C.$$ |
$$x = \ln\sqrt{\left|\frac{1\!+\!u}{1\!-\!u}\right|}+C, \quad y = \ln\sqrt{\left|\frac{1\!+\!u}{1\!-\!u}\right|}-u\!+\!C.$$ |
|
|
| \begin{thebibliography}{9} |
\begin{thebibliography}{9} |
|
\bibitem{3L}{\sc E. Lindel\"of:} {\em Differentiali- ja integralilasku III 1}.\, Mercatorin Kirjapaino Osakeyhti\"o, Helsinki (1935).
|
\bibitem{3L}{\sc E. Lindel\"of:} {\em Differentiali- ja integralilasku} III 1.\, Mercatorin Kirjapaino Osakeyhti\"o, Helsinki (1935).
|
| \end{thebibliography} |
\end{thebibliography} |
