Differential equations

The following notations for derivatives are used: and ${y}''\to \frac{{{d}^{2}}y}{d{{x}^{2}}}$, that is .

Definition. A differential equation is a relation between an independent variable x, the unknown function and its derivatives $y',{y}'',{y}''',...,{{y}^{(n)}}$, of the form

If the unknown function $y$ is a function of a single variable x, the differential equation is called ordinary.

Definition. A solution of a differential equation on an interval (a,b) is a function $y=\varphi (x)$ defined on that interval together with its derivatives, and for which substituting $y=\varphi (x)$ into the differential equation turns it into an identity in x on (a,b).

To determine all the functions which are solutions of a differential equation is to solve that differential equation.

A separable differential equation has the form: where $F:I\to \mathbb{R}$ and $g:J\to \mathbb{R}$ are two continuous functions. The equation is equivalent to $g(y)dy=f(x)dx$. Integrating, we obtain $(G\circ y)(x)=F(x)+c$ and $y(x)={{G}^{-1}}(F(x)+c),$ where G is an antiderivative of g, F is an antiderivative of f, and $c\in \mathbb{R}$.

A first-order linear differential equation has the form: $y'=p(x)*y+q(x),$ where $p,q:I\to \mathbb{R}$ are continuous functions. The solution of this equation is a differentiable function $\varphi :I\to \mathbb{R}$ satisfying: $\varphi '(x)=p(x)\cdot \varphi (x)+q(x), \forall x\in I.$

First-order differential equations: , where p, t are continuous functions on an interval (a,b).

Every solution of the first-order ordinary equation is the sum of a particular solution of that equation and the general solution of the equation

is determined for constant and for a particular form of , as shown in the table below.

| | | | Form of $t(x)$ | Form of the particular solution | | Polynomial | Polynomial of the same degree | | $a{{e}^{rx}}$ | $\alpha {{e}^{rx}}$ | | $a{{e}^{-rx}}$ | $\alpha x{{e}^{-rx}}$ | | $a\cos (rx)+b\sin (rx)$ | |

Solving the homogeneous equation:$y'+p(x)\cdot y=0$

The homogeneous equation $y'+p(x)\cdot y=0$, $p:I\to \mathbb{R}$ has the solution $\varphi :I\to \mathbb{R},$ $\displaystyle \varphi (x)=k\cdot {{e}^{-\int\limits_{{{x}_{0}}}^{x}{p(t)dt}}},$ $\forall x\in I$.

The second-order differential equation has the form $F(x,y,y',{y}'')=0.$

Equations of the form $ay''+by'+cy=h(x)$ are called second-order differential equations with constant coefficients.

The second-order linear homogeneous differential equation

The equation with ${{h}^{2}}-4k>0, {{x}_{0}}{{y}_{0}},y{{'}_{0}}\in \mathbb{R}$ has at least one solution $f:\mathbb{R}\to \mathbb{R}$ of the form $f(x)=a\cdot {{y}_{1}}(x)+b\cdot {{y}_{2}}(x)$. The solutions of the characteristic equation, ${{r}^{2}}+hr+k=0$ are ${{r}_{1}}, {{r}_{2}}$, and a and b are real numbers determined from the conditions and $f'({{x}_{0}})=y{{'}_{0}}.$

We have the following cases:

$\Delta$	solutions of the equation ${{r}^{2}}+hr+k=0$	solution of the equation $y''+hy'+ky=0$
$\begin{aligned} & {{h}^{2}}-4k>0 \\ & {{h}^{2}}-4k=0 \\ & {{h}^{2}}-4k<0 \\ \end{aligned}$	$\begin{aligned} & {{r}_{1}},{{r}_{2}},reale,distincte \\ & {{r}_{1}}={{r}_{2}} \\ & {{r}_{1}}=\overline{{{r}_{2}}}\in \mathbb{C}\backslash \mathbb{R},{{r}_{1}}=\alpha +i\beta \\ \end{aligned}$	$\begin{aligned} & f(x)=a{{e}^{{{r}_{1}}x}}+b{{e}^{{{r}_{2}}x}} \\ & f(x)=(ax+b){{e}^{{{r}_{1}}x}} \\ & f(x)={{e}^{\alpha x}}(a\cos (\beta x)+b\sin (\beta x)) \\ \end{aligned}$

Graphical representation of the solutions of differential equations

1. A body heated to ${{160}^{\circ }}C$ is immersed in a medium held at a constant temperature of . Within one minute the temperature of the body falls to . Let be the temperature of the body at time t and ${{T}_{0}}$ the constant temperature of the medium. Then satisfies the differential equation: ${T}'(t)=k\cdot (60-T(t))$, that is ${T}'(t)=-k\cdot T(t)+60\cdot k$, which has the analytic solution: . From the initial conditions and .

From these conditions we obtain: $c=100, k=\ln \left( \frac{5}{2} \right)$. Therefore $T(t)=60+100\cdot {{e}^{-t\ln \left( \frac{5}{2} \right)}}.$