First-order ODE

Definition

Ordinary Differential Equation of order 1

implicit form

$F(x,y,y^{\prime })=0$

explicit form

$y^{\prime }=G(x,y)$

Facts

we say $y=f(x)$ is a solution on some open interval $a<x<b$ if

• $f$ is defined and differentiable on $(a,b)$
• $F(x,f(x),f^{\prime }(x))=0$ for $x\in (a,b)$

The graph of $f$ is a solution curve

general solution

A solution containing an arbitrary contant $C$

If we choose a specific $C$ for general solution, then we obtain a particular solution

Solutions

Separable ODE

Definition

$g(y)y^{\prime }=f(x)$

$g(y)dy=f(x)dx$

Solution

can be solved by integration $\int g(y)dy=\int f(x)dx+C$

Reduction to Separable ODE

If there is an ODE of the form $y^{\prime }=f(\frac{y}{x})$, then let $\frac{y}{x}=u$. Then $y=xu\Rightarrow y^{\prime }=u^{\prime }x+u\Rightarrow u^{\prime }x+u=f(u)\Rightarrow \frac{du}{f(u)-u}=\frac{dx}{x}$:

Examples

&\cfrac{dy}{dx} = (y^{2} + 1)\\ &\cfrac{1}{y^{2}+1}dy = dx\\ &\int \cfrac{1}{y^{2}+1}dy = \int 1 dx = tan^{-1}(y) = x+C\\ &y = \tan(x+C) \end{aligned}$$ $$\begin{aligned} &2xy\frac{dy}{dx} = y^{2}-x^{2}\\ & \frac{dy}{dx} = \frac{1}{2}\left( \frac{y}{x} - \frac{x}{y} \right)\\ &\text{substitute}\ u:=\frac{y}{x} \Rightarrow u+x\frac{du}{dx} = \frac{dy}{dx}\\ & u+x\frac{du}{dx} = \frac{1}{2}\left( \frac{u^{2}-1}{u} \right)\\ & x\frac{du}{dx} = -\frac{1}{2}\left( \frac{u^{2}+1}{u} \right)\\ & \int\left( \frac{2u}{u^{2}+1} \right)du = -\int\frac{1}{x}dx\\ & \ln|u^{2}+1| = -\ln|x|+C\\ & |u^{2}+1| = \frac{1}{x}e^{C}\\ & u^{2}+1 = \pm \frac{1}{x}e^{C}\\ & \frac{y^{2}}{x^{2}} = \frac{1}{x}C -1\\ & y = \pm \sqrt{Cx - x^{2}} \end{aligned}$$Link to original

Linear ODE

Definition

A first-order ODE that can be brought into standard form $y^{\prime }+p(x)y=r(x)$:

Solution

Transclude of Homogeneous-Linear-ODE

Transclude of Non-Homegeneous-Linear-ODE

Transclude of Bernoulli-Equation

Exact ODE

Definition

A first-order ODE $M(x,y)+N(x,y)y^{\prime }=0$, written as $Mdx+Ndy=0$ is called an exact ODE if $Mdx+Ndy=0$ is exact.

Solution

If $Mdx+Ndy=0$ is exact, $Mdx+Ndy=du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$ (for some $u(x,y)$) $\Rightarrow du=0$ and $u(x,y)=c$

If we only know $\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\frac{\partial u}{\partial y}=\frac{\partial N}{\partial x}$ ^[by Clairaut’s Theorem $f_{x}y(a,b)=f_{y}x(a,b)$] $M=\frac{\partial u}{\partial x}\to u=\int Mdx+k(y)$ $N=\frac{\partial u}{\partial y}\to u=\int Ndy+l(y)$ where $k$ is determined by $\frac{\partial u}{\partial y}$ and $l$ is determined by $\frac{\partial u}{\partial x}$

Finding $k(y)$ or $l(x)$

Differentiate $u=\int Mdx+k(y)$ with respect to $y$ $N=\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}(\int Mdx+k(y))$ $\Rightarrow \frac{\partial }{\partial y}k(y)=N-\frac{\partial }{\partial y}(\int Mdx)$ $\Rightarrow k(y)=\int (N-\frac{\partial }{\partial y}(\int Mdx))\partial y$

Substitute $k(x)$ in the original expression $u=\int Mdx+k(y)$

$u=\int Mdx+\int (N-\frac{\partial }{\partial y}(\int Mdx))\partial y$

or, similar to above, find $l(x)$ in the expression below $N=\frac{\partial u}{\partial y}\to u=\int Ndy+l(x)$ (where $l$ is determined by $\frac{\partial u}{\partial x}$)

Examples

&\left( \frac{y}{x} + 4x \right)dx + \left( \ln x - 3 \right)dy = 0\\ &\frac{\partial \left( \frac{y}{x} + 4x \right)}{\partial y} = \frac{\partial \left( \ln x - 3 \right)}{\partial y} = \frac{1}{x}: \text{Exact ODE}\\ &\int\left( \frac{y}{x} + 4x \right)dx = y\ln|x|+2x^{2} + h(y) = f\\ &f_{y} = \ln|x|+h'(y) = \ln|x|-3 \Rightarrow h(y) = -3y + C\\ &f = y\ln|x|+2x^{2} - 3y = C \end{aligned}$$ # Facts > If $u(x, y)$ can be expressed the form of $du = Mdx + Ndy$^[result of [[Total differntial]]], $u$ is exact ^632ad5 > $Mdx + Ndy = du =0$ is exact $\Leftrightarrow \frac{\partial M}{\partial y} + \frac{\partial N}{\partial x} = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy$ (i.e. $M_y = N_x$)Link to original

Integrating Factors

Definition

Let $Pdx+Qdy=0$ be an ODE.

If there exist a function $F$ s.t. $FPdx+FQdy=0$, then $F(x,y)$ is called an integrating factor

Finding

$Pdx+Qdy=0\Rightarrow FPdx+FQdy=0\Rightarrow \frac{\partial }{\partial y}FP=\frac{\partial }{\partial x}FQ\Rightarrow F_{y}P+F{P}_y=F_{x}Q+F{Q}_x$

If $F=F(x)$ and $F_y=0$, then $F{P}_y=F_{x}Q+F{Q}_x\Rightarrow \frac{dF}{dx}Q=F(P_y-Q_x)\Rightarrow \frac{dF}{dx}\frac{1}{F}=\frac{1}{Q}(P_y-Q_x)=R(x)$. If $R$ depends only on $x$, $\ln |F|=\int Rdx+C\Rightarrow F(x)=\exp (\int R(x)dx)$

If $F=F(y)$ and $F_x=0$, similarly, $\frac{1}{P}(Q_x-P_y)=R^{\ast }(y)$. If $R^{\ast }$ depends only on $y$, $F^{\ast }(y)=\exp (\int R^{\ast }(y)dy)$

Examples

&e^{x} + (e^{x}\cot(y) + 2y \csc(y))y' = 0\\ &\underbrace{e^{x}}_{P}dx + \underbrace{(e^{x}\cot(y) + 2y \csc(y))}_{Q}dy = 0\\ &P_{y}=0,\quad Q_{x}=e^{x}\cot(y), \quad R(x) = \frac{Q_{x}-P_{y}}{P}=\cot(y), F(y)=\exp(\int\cot(y)dy)=\sin(y)\\ &\underbrace{\sin(y)e^{x}}_{P'}dx + \underbrace{(e^{x}\cos(y) + 2y)}_{Q'}dy = 0\\ &P'_{y} = Q'_{x} = \cos(y)e^{x}: \text{Exact ODE}\\ &\int p' dx = \sin(y)e^{x} + h(y) = f\\ &f_{y} = e^{x}\cos(y) + h'(y) = (e^{x}\cos(y) + 2y) \Rightarrow h'(y) = 2y \Rightarrow h(y)=y^{2}\\ &f = \sin(y)e^{x} + y^{2}= C \end{aligned}$$Link to original

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