Differential-equation Formula Sheet

A differential equation is a mathematical equation that relates a function to its derivatives, representing how the function changes over time or space.
It involves an unknown function and its derivatives, and it describes the relationship between the rates of change of the function and the function itself.

Neetesh Kumar | June 03, 2024 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition
• 2. Solution (Primitive) of Differential Equation
• 3. Order of Differential Equation
• 4. Degree of Differential Equation
• 5. Formation of a Differential Equation
• 6. General and Particular Solutions
• 7. Elementary Types of First Order & First Degree Differential Equations
• 8. Linear Differential Equations
• 9. Trajectories

1. Definition:

An equation that involves independent and dependent variables and the derivatives of the dependent variables is called a differential equation.

2. Solution (Primitive) of Differential Equation:

Finding the unknown function which satisfies the given differential equation is called solving or integrating the differential equation. The solution of the differential equation is also called its primitive because the differential equation can be regarded as a relation derived from it.

3. Order of Differential Equation:

The order of a differential equation is the order of the highest-order differential coefficient occurring in it.

4. Degree of Differential Equation:

The degree of a differential equation, which can be written as a polynomial in the derivatives, is the degree of the highest order derivative occurring in it after it has been expressed in a form free from radicals and fractions so far as derivatives are concerned.

Thus, the differential equation:

$f(x, y) \left[\frac{d^m y}{dx^m}\right]^p + \phi(x, y) \left[\frac{d^{m-1} \left(y\right)}{dx^{m-1}}\right]^q + \ldots = 0$ is of order $m$ and degree $p$.

Note that in the differential equation $e^{y'''} - xy'' + y = 0$, the order is three but the degree doesn’t exist.

5. Formation of a Differential Equation:

If an equation in independent and dependent variables having some arbitrary constants is given, then a differential equation is obtained as follows:

(a) Differentiate the given equation concerning the independent variable (say $x$) as many times as the number of independent arbitrary constants.

(b) Eliminate the arbitrary constants.

The eliminant is the required differential equation.

Note: A differential equation represents a family of curves all satisfying some common properties. This can be considered as the geometrical interpretation of the differential equation.

6. General and Particular Solutions:

The solution of a differential equation that contains several independent arbitrary constants equal to the order of the differential equation is called the general solution (or complete integral or complete primitive).
A solution is obtainable from the general solution by giving particular or initial values to the constants, which is called a particular solution.

7. Elementary Types of First Order & First Degree Differential Equations:

(a) Variables separable:

Type-1: If the differential equation can be expressed as: $f(x)dx + g(y)dy = 0$ then this is said to be a variable-separable type. A general solution of this is given by: $\int f(x) \, dx + \int g(y) \, dy = c$ where $c$ is the arbitrary constant. Consider the example $\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}$.

Type-2: Sometimes, transformation to polar coordinates facilitates the separation of variables. In this connection, it is convenient to remember the following differentials. If $x = r \cos \theta$, $y = r \sin \theta$, then:

1. $xdx + ydy = rdr$
2. $dx^2 + dy^2 = dr^2 + r^2 d\theta^2$
3. $xdy - ydx = r^2 d\theta$

Also, if $x = r \sec \theta$ & $y = r \tan \theta$ then:

1. $xdx - ydy = rdr$
2. $xdy - ydx = r^2 \sec \theta d\theta$

Type-3:

$\frac{dy}{dx} = f(ax + by + c), \quad b \neq 0$ To solve this, substitute $t = ax + by + c$.
Then, the equation is reduced to a separable type in the variables $t$ and $x$, which can be solved. Consider the example: $\frac{dy}{dx} = \frac{x^2 + y^2 + a}{x^2 - y^2}$

(b) Homogeneous equations:
A differential equation of the form $\frac{dy}{dx} = \frac{f(x,y)}{g(x,y)}$ where $f(x,y)$ and $g(x,y)$ are homogeneous functions of $x$ and $y$ and of the same degree, is called homogeneous.
This equation may also be reduced to the form $\frac{dy}{dx} = \frac{g(x)}{h(y)}$ and is solved by putting $y = vx$ so that the dependent variable $y$ is changed to another variable $v$, where $v$ is some unknown function.
Thus, the differential equation is transformed into an equation where the variables are separable. Consider the example: $\frac{dy}{dx} = \frac{x + y}{x - y}$

(c) Equations reducible to the homogeneous form:
If $\frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}$ where $a_1 b_2 - a_2 b_1 \neq 0$, i.e. $a_1 / a_2 \neq b_1 / b_2$, then the substitution $x = u + h$, $y = v + k$ transforms this equation to a homogeneous type in the new variables $u$ and $v$ where $h$ and $k$ are arbitrary constants to be chosen to make the given equation homogeneous.

Note:
1. If $a_1 b_2 - a_2 b_1 = 0$, then a substitution $u = a_1 x + b_1 y$ transforms the differential equation to an equation with variables separable.

2. If $b_1 + a_2 = 0$, then a simple cross multiplication and substituting $d(xy)$ for $xdy + ydx$ and integrating term by term yields the result easily.
Consider the examples: $\frac{dy}{dx} = \frac{2x + 3y - 1}{4x - 6y + 5}$ $\frac{dy}{dx} = \frac{2x - y + 1}{6x + 5y - 4}$

3. In an equation of the form:

$yf(xy)dx + xg(xy)dy = 0$ the variables can be separated by the substitution $xy = v$.

8. Linear Differential Equations:

A differential equation is said to be linear if the dependent variable and its differential coefficients occur in the first degree only and are not multiplied together.

The $n$th order linear differential equation is of the form:
$a_0(x) \frac{d^n y}{dx^n} + a_1(x) \frac{d^{n-1} y}{dx^{n-1}} + \ldots + a_{n-1}(x) \frac{dy}{dx} + a_n(x)y = 0$
where $a_0(x), a_1(x), \ldots, a_n(x)$ are called the coefficients of the differential equation.

(a) Linear differential equations of first order:
The most general form of linear differential equations of first order is:
$\frac{dy}{dx} + Py = Q$ where $P$ and $Q$ are functions of $x$.

To solve such an equation, multiply both sides by the integrating factor:
$e^{\int P \, dx}$ Then the solution of this equation will be: $y e^{\int P \, dx} = \int Q e^{\int P \, dx} \, dx + c$

(b) Equations reducible to linear form:
The equation:$\frac{dy}{dx} + Py = Q y^n$

where $P$ and $Q$ are functions of $x$, is reducible to the linear form by dividing it by $y^n$ and then substituting $y^{-n+1} = Z$. Consider the example:

$(x^3 y^2 + xy) dx = dy$

The equation: $\frac{dy}{dx} + Py = Q y^n$ is called Bernoulli's equation.

9. Trajectories:

A curve which cuts every member of a given family of curves according to a given law is called a trajectory of the given family.

Orthogonal trajectories:
A curve making at each of its points a right angle with the curve of the family passing through that point is called an orthogonal trajectory of that family.

We set up the differential equation of the given family of curves. Let it be $F(x, y, y') = 0$.
The differential equation of the orthogonal trajectories is of the form: $F \left( x, y, \frac{1}{y'} \right) = 0$
The general integral of this equation $f(x, y, C) = 0$ gives the family of orthogonal trajectories.

Note: The following exact differentials must be remembered:

(i) $dx + dy = d(x + y)$

(ii) $dx - dy = d(x - y)$

(iii) $x \, dy + y \, dx = d(xy)$

(iv) $\frac{x \, dy - y \, dx}{x^2} = y \, d \left( \frac{x}{y} \right)$

(v) $\frac{y \, dx - x \, dy}{y^2} = x \, d \left( \frac{y}{x} \right)$

(vi) $2(x \, dx + y \, dy) = d(x^2 + y^2)$

It should be observed that:

(i) $\frac{\left( x \, dy + y \, dx \right)}{xy} = d (\ln xy)$

(ii) $\frac{\left( dx + dy \right)}{x+y} = d \left( \ln \left( \frac{x}{y} \right) \right)$

(iii) $\frac{x \, dy - y \, dx}{xy} = y \, d \left( \ln \left( \frac{x}{y} \right) \right)$

(iv) $\frac{y \, dx - x \, dy}{xy} = x \, d \left( \ln \left( \frac{y}{x} \right) \right)$

(v) $\frac{x \, dy - y \, dx}{x^2 + y^2} = d \left( \tan^{-1} \left( \frac{y}{x} \right) \right)$

(vi) $\frac{y \, dx - x \, dy}{x^2 + y^2} = d \left( \tan^{-1} \left( \frac{x}{y} \right) \right)$

(vii) $\frac{xdx + ydy}{x^2 + y^2} = d (\ln (x^2 + y^2))$

(viii) $\frac{x \, dy + y \, dx}{xy} = d \left( \frac{x}{y} \right)$

(ix) $x e^x \, dx + e^y \, dy = d \left( \frac{e^x}{e^y} \right)$

(x) $y e^y \, dy + e^x \, dx = d \left( \frac{e^y}{e^x} \right)$

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