Differential Equations!

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Def. 1 - Order

If $y^{(n)}$ is the highest order derivative of $y$ that appears in (1), then we say (1) is the $n$ th order ODE.

In other words, an equation has order $n$ if this equation contains $y^{(n)}$ , and doesn’t contain any higher order derivative of $y$ .

Example (Example 1)

$y'' + 2y = 0$ — this is a 2nd order ODE.

Example (Example 2)

$y'''y + (y'')^2 = x^2$ — this is a 3rd order ODE.

Def. 2 - Linear vs Nonlinear

If an equation has the form

a_n(x)\cdot y^{(n)}(x) + a_{n-1}(x)\cdot y^{(n-1)}(x) + \cdots + a_2(x)\cdot y^{(2)}(x) + a_1(x)\cdot y'(x) + a_0(x)\cdot y(x) + a(x) = 0

here $a_n(x), a_{n-1}(x), \ldots, a_1(x), a_0(x)$ are functions of $x$ (functions we already know), then we say this is a linear ODE. Otherwise, this is a non-linear ODE.

Example (Example 3)

$x^2\cdot y'' + \sin(x)\cdot y' + x^3 = 0$ — this is a 2nd order linear ODE.

If we let $a_2(x) = x^2$ , $a_1(x) = \sin(x)$ , $a_0(x) = 0$ , $a(x) = x^3$ , then (2) has the form

a_2(x)\cdot y^{(2)}(x) + a_1(x)\cdot y^{(1)}(x) + a_0(x)\cdot y(x) + a(x) = 0.

Example (Example 4)

$y'(x) + y^2(x) = x^2$ — this is a 1st order non-linear ODE, because we can’t write it in the form of (2) due to the existence of $y^2$ .

Example (Example 5)

$y''(x)\cdot y(x) + y'(x) = x$ — this is a 2nd order non-linear ODE due to the existence of $y''(x)\cdot y(x)$ .

In a linear equation, you can always separate $y(x)$ , $y'(x)$ , $y''(x)$ (no products like $y''(x)\cdot y(x)$ ).

Def. 3 - Solution

If $y=y(x)$ satisfies (solves)

F\bigl(x, y(x), y'(x), y''(x)\bigr) = 0 \quad (1)

then we say $y(x)$ is a solution of (1).

First-order linear ODE

\frac{dy}{dx} + P(x)\cdot y = Q(x)

here $P(x), Q(x)$ are functions of $x$ . This is a 1st order linear ODE.

If $a_1(x) = 1$ , $a_0(x) = P(x)$ , and $a(x) = -Q(x)$ , then (3) can be written as

a_1(x)\cdot y^{(1)}(x) + a_0(x)\cdot y(x) + a(x) = 0.

\frac{dy}{dx} + 2\cdot y(x) = 0 \quad (4)

We times $e^{2x}$ on both sides of (4), then we have

e^{2x}\cdot \frac{dy}{dx} + 2\cdot e^{2x}\cdot y(x) = 0 \quad (5)

From (5), we obtain

\bigl(e^{2x}\cdot y(x)\bigr)' = 0.

\bigl(e^{2x}\cdot y(x)\bigr)' = (e^{2x})'\cdot y(x) + e^{2x}\cdot y'(x) = 2\cdot e^{2x}\cdot y(x) + e^{2x}\cdot y'(x).

Since if $f' = 0$ , then $f=C$ (a constant), from the above we have

e^{2x}\cdot y(x) = C, \quad y(x) = \frac{C}{e^{2x}} = C\cdot e^{-2x},

here $C$ can be any constant.

Plotting one solution curve for $y = C\cdot e^{-2x}$ (take $C = 1$ ):

Slope field for the differential equation $y' = y - x$ :

3D surface for $z = \sin(x)\cos(y)$ :