First-order differential equations are mathematical equations that describe how one quantity depends on another quantity. They can be used to describe many phenomena in nature, such as population growth, battery discharge, or fluid flow.
A first-order differential equation can be written as: dy/dx = f(x, y), where x and y are independent and dependent variables respectively, and f(x, y) is a function that describes the relation between x and y.

Differential equations (Diff. Eq.) with separable variables 1

Form : $(\varphi (y){)}^{\prime }=f(x)\to$Solution (Primitives of the 2 members): $\varphi (y)=F(x)+C$ with $C$ a constant
Example:
Solve Diff. Eq. $(E):3.x.y^{\prime }=y$ where $y$ a nonzero function and $x\in ]0;+\mathrm{\infty }[$
Solution: The equation $(E)$ is written: $\frac{y^{\prime }}{y}=\frac{1}{3x}$
We obtain: $ln|y|=\frac{1}{3}ln|x|+C$ which is written: $ln|y|=\frac{1}{3}ln(x)+C$ for $x\in ]0;+\mathrm{\infty }[$
$|y|=e^{ln(x\frac{1}{3}+C)}=x^{\frac{1}{3}}\ast e^C=\alpha \sqrt[3]{x}$ with $\alpha$ real constant < 0
We can summarize all the solutions by the single notation: $y=k.\sqrt[3]{x}$, with $k\in \mathbb{R}$

Linear differential equations of order 1

Definition:
Linear: a function and its derivative occur separately
Form : $(E):a(x).y^{\prime }(x)+b(x).y(x)=f(x)$ where $a$,$b$ and $f$ functions of the variable $x$
“With constant coefficients” means that $a$ and $b$ are constant $\to (E):a.y^{\prime }(x)+b.y(x)=f(x)$
Second member: function $f$ is the second member, if $f(x)=0$ the equation has no second member.

Equation without 2nd member

Let $(E_0):a.y^{\prime }(x)+b.y(x)=0$ with $a(x)\ne 0$
All the solutions of $(E_0)$ on the interval $I$ are the functions of the form: $y_0(x)=k.e^{-G(x)}$ where $G$ is the antiderivative of the function $g(x)=\frac{b(x)}{a(x)}$ and $k$ a real constant
$\to$ We say that $y_0(x)=k.e^{-G(x)}$ is the general solution of $(E_0)$ on the interval $I$.< br>

Equation with 2nd member

$y_p(x)$ is a particular solution of $(E)$.
$\to$ All the solutions of $(E)$ on the interval $I$ are the sum of the general solution and the particular solution: $y(x)=y_0(x)+y_p(x)$

Search for a particular solution: Constant variation method

Let $y_0(x)=k.e^{-G(x)}$ the solutions are different from 0 of $(E_0)$. We are looking for a particular solution $y_p(x)$ of the same type as $y_0(x)$ by “varying the constant”.
We set: $y_p(x)=k(x).{\varphi }_0(x)$\) where ${\varphi }_0(x)=e^{-G(x)}$ and $k$ a function to be determined.
All the solutions of $(E)$ on the interval $I$ are the shape functions: $y(x)=y_0(x)+k(x).{\varphi }_0(x)$
Where the function $k$ is primitive of the function:
$$h(x)=\frac{f(x)}{a(x).{\varphi }_0(x)}$$ Example:
Let Diff. Eq : $$(E):y^{\prime }(x)+3.y(x)=x$$ We note $y_0(x)$ the general solution of Diff. Eq. without 2nd member: $(E_0):y^{\prime }(x)+3.y(x)=0$
$y_0(x)=k.e^{-3x}$ where $k$ a real constant
To determine the particular solution, we apply the method of variation of the constant:
$$h(x)=\frac{f(x)}{a(x).{\varphi }_0(x)}=\frac{x}{e^{-3x}}=x.e^{3x}$$ $$y_p(x)=k(x).{\varphi }_0(x)\text{ with }{k}^{\prime }(x)=h(x)=x.e^{3x}$$ An integration by parts allows to obtain the expression of the function $k$:
$$k(x)=\frac{1}{3}.x.e^{3x}-\frac{1}{9}.e^{3x}$$ We deduce the general solution: $$y(x)=y_0(x)+k(x).{\varphi }_0(x)=k{e}^{-3x}+(\frac{1}{3}.x.e^{3x}-\frac{1}{9}.e^{3x}).e^{-3x}$$ $$y(x)=k.e^{-3x}+\frac{1}{3}.x-\frac{1}{9}\text{ where k is a real constant}$$

Diff. Eq. of order 1 with constant coefficients

We can look for a particular solution of the same type as the 2nd member $f(x)$.
This method works for functions such as polynomials, sinusoids or exponentials.
Example
Let Diff. Eq. :$(E)y^{\prime }(x)+3.y(x)=x$
We note $y_0(x)$ the general solution of Eq. without 2nd member:$(E_0):y^{\prime }(x)+3.y(x)=0$
$y_0(x)=k.e^{-3x}$ where $k$ a real constant
To determine the particular solution, we seek a solution of the form: $y_p(x)=ax+b$
We calculate the derivative: $y_p^{\prime }(x)=a$ and we replace by $(E):a+(ax+b)\ast 3=x$
We get the following system:
$$\{\begin{array}{l}3.a=1\\ a+3.b=0\end{array}\leftrightarrow \{\begin{array}{l}a=\frac{1}{3}\\ b=-\frac{1}{9}\end{array}$$ Finally:
$$y(x)=k.e^{-3x}+\frac{1}{3}.x-\frac{1}{9}$$ where $k$ a real constant

Differential equations (Diff. Eq.) with separable variables 2

Can be written:
$$a(y).y^{\prime }=b(x)\leftrightarrow a(y).\frac{dy}{dx}=b(x)\leftrightarrow a(y).dy=b(x).dx$$ with $a$,$b$ 2 functions

Resolution

It is necessary to primitiveize the 2 members of the equality: $A(y)=B(x)+C$ with $C$ a real constant, $A$ a primitive of $a$, $B$ a primitive of $b$
The solutions $a(y).y^{\prime }=b(x)$ are $y=f(x)$ which satisfy $A(y)=B(x)+C$
If we can invert $A$, with general solution: $f(x)=A^{-1}(B(x)+C)$
To determine the particular solution of Diff. Eq. checking $(x_0;y_0)$:
Find the general solution then determine $C$ with $y_0=f(x_0)$ (1)
Or, integrate the equation in separate form from the initial condition (2):
The solution of $a(y).y^{\prime }=b(x)$ satisfying the initial condition $(x_0,y_0)$ is the function $y=f(x)$ such that:
$${\int }_{y_0}^{y}a(u)du={\int }_{x_0}^{x}b(t)dt$$ $(1)\to$ long method

$(2)\to$ short method
Example:
Determine the solution of Diff. Eq. :
$y^{\prime }=4.x^3(1+y)$ which satisfies the initial condition $(x_0;y_0)=(1;0)$
$y^{\prime }=4x^3(1+y)\leftrightarrow \frac{y^{\prime }}{1+y}=4.x^3$ or $\frac{dy}{1+y}=4.x^{3}dx$ (we put all the identical variables together)
We integrate: $\frac{dy}{1+y}=4.x^{3}dx$ from the bounds corresponding to the initial condition:
$${\int }_0^y\frac{du}{1+u}={\int }_1^{x}4t^{3}dt\leftrightarrow [ln(u+1){]}_0^y=[t^4{]}_1^x\leftrightarrow ln(y+1)-0=x^4-1\leftrightarrow y+1=e^{x^4-1}\leftrightarrow y=-1+e^{x^4-1}$$

Linear differential equations of order 1

Form : $(E):a(x).y^{\prime }(x)+b(x).y(x)=f(x)$ where $a$,$b$ and $f$ functions of the variable $x$
If $c(x)=0$, then the equation is without 2nd member
Principle of resolution:
1) Solve Diff. Eq. without 2nd member: homogeneous equation
$$(H):a(x).y^{\prime }(x)+b(x).y(x)=0$$ By this:
$$a(x).y^{\prime }(x)+b(x).y(x)=0\to y^{\prime }(x)=-\frac{b(x)}{a(x)}.y(x)=u(x).y(x)\text{ where }u(x)=-\frac{b(x)}{a(x)}$$ Let $U$ be a primitive of $u$: $y=\lambda .e^{U(x)}$, with $\int -\frac{b(x)}{a(x)}dx=U(x)+C$
After solving, we obtain the general solution in the form: $y=\lambda .g(x)$ with $g(x)=e^{U(x)}$
Example:
$$I=]-\frac{\pi }{2};\frac{\pi }{2}[,(cos(x)).y^{\prime }+(sin(x)).y=0\to y^{\prime }=-\frac{sin(x)}{cos(x)}.y$$ We have:
$$\int -\frac{sin(x)}{cos(x)}dx=ln|cos(x)|+C$$ So the general solution is:
$$y=\lambda .e^{ln|cos(x)|}=\lambda .cos(x)$$ 2) Find the particular solution
2 methods:
– By intuition, often the case when the 2nd member is constant or a simple polynomial. We seek the particular solution by identification.
– Using the constant variation method:
We replace the constant \lambda by a function $\lambda (x)$: $y=\lambda (x).g(x)$
$$y^{\prime }={\lambda }^{\prime }(x).g(x)+\lambda (x).g^{\prime }(x)\to \text{ Solution when }a(x).y^{\prime }+b(x).y=c(x)$$ $$a(x).({\lambda }^{\prime }(x).g(x)+\lambda (x).g^{\prime }(x))+b(x).(\lambda (x).g(x))=c(x)$$ $$\leftrightarrow a(x).{\lambda }^{\prime }(x).g(x)+\lambda (x).(a(x).g^{\prime }(x)+b(x).g(x))=c(x)$$ Now $g$ solution of $H$, therefore: $a(x).g^{\prime }(x)+b(x).g(x)=0$, the solutions of $(E)$ are the functions:
$$y=\lambda (x).g(x)\text{ with }{\lambda }^{\prime }(x)=\frac{c(x)}{)}a(x).g(x)$$ Doing the primitive :
$$\int \frac{c(x)}{a(x).g(x)}=F(x)+C$$ The solutions of $(E)$ are:
$$y=(F(x)+C).g(x)=F(x).g(x)+C.g(x)$$ Example:
Solve: $(E):2.x.y^{\prime }–y=\sqrt{x}$ on $I=]0;+\mathrm{\infty }[$ $$(H):2.x.y^{\prime }-y=0\leftrightarrow y^{\prime }=\frac{1}{2x}.y$$ $$\int \frac{1}{2x}.ydx=\frac{1}{2}.ln|x|+C=ln(x^{\frac{1}{2}})+C$$ $$y=\lambda .e^{ln|x^{\frac{1}{2}}|}=\lambda .\sqrt{x}$$ The solutions of $(E)$:
$$2x.{\lambda }^{\prime }(x).\sqrt{x}+\lambda (x).(2x.g^{\prime }(x))-g(x)=\sqrt{x}\leftrightarrow {\lambda }^{\prime }(x)=\frac{1}{2x}$$ $$(2x.g^{\prime }(x))-g(x)=0\text{ because g solution of H}$$ $$\lambda (x)=\frac{1}{2}.ln(x)+C$$ The general solution of $(E)$ is:
$$y=(\frac{1}{2}.ln(x)+C).\sqrt{x}=\sqrt{x}.ln(\sqrt{x})+C.\sqrt{x}$$

Tagged: differential equations; first order; flow of a fluid; resolution