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How To Find The Order Of Differential Equation And Its Degree?

Order of Differential Equation:-

Differential Equations are classified on the basis of the order. Order of a differential equation is the order of the highest derivative (also known as differential coefficient) present in the equation.

Example (i):

\(\begin{array}{l}\frac{d^3 x}{dx^3} + 3x\frac{dy}{dx} = e^y\end{array} \)

In this equation, the order of the highest derivative is 3 hence, this is a third order differential equation.

Example (ii) : –

\(\begin{array}{l} (\frac{d^2 y}{dx^2})^ 4 + \frac{dy}{dx}= 3 \end{array} \)

This equation represents a second order differential equation.

This way we can have higher order differential equations i.e.

\(\begin{array}{l} n^{th}\end{array} \)
order differential equations.

Differential-Equation

First order differential equation:

The order of highest derivative in case of first order differential equations is 1. A linear differential equation has order 1. In case of linear differential equations, the first derivative is the highest order derivative.

\(\begin{array}{l}\frac{dy}{dx} + Py = Q \end{array} \)

P and Q are either constants or functions of the independent variable only.

This represents a linear differential equation whose order is 1.

Example:

\(\begin{array}{l} \frac{dy}{dx} + (x^2 + 5)y = \frac{x}{5} \end{array} \)

This also represents a First order Differential Equation.

Second Order Differential Equation:

When the order of the highest derivative present is 2, then it is a second order differential equation.

Example:

\(\begin{array}{l}\frac{d^2 y}{dx^2} + (x^3 + 3x) y = 9 \end{array} \)

In this example, the order of the highest derivative is 2. Therefore, it is a second order differential equation.

Degree of Differential Equation:

The degree of the differential equation is represented by the power of the highest order derivative in the given differential equation.

The differential equation must be a polynomial equation in derivatives for the degree to be defined.

Example 1:-

\(\begin{array}{l}\frac{d^4 y}{dx^4} + (\frac{d^2 y}{dx^2})^2 – 3\frac{dy}{dx} + y = 9 \end{array} \)

Here, the exponent of the highest order derivative is one and the given differential equation is a polynomial equation in derivatives. Hence, the degree of this equation is 1.

Example 2:

\(\begin{array}{l} [\frac{d^2 y}{dx^2} + (\frac{dy}{dx})^2]^4 = k^2(\frac{d^3 y}{dx^3})^2\end{array} \)

The order of this equation is 3 and the degree is 2 as the highest derivative is of order 3 and the exponent raised to the highest derivative is 2.

When the Degree of Differential Equation is not Defined?

It is not possible every time that we can find the degree of given differential equation. The degree of any differential equation can be found when it is in the form a polynomial; otherwise, the degree cannot be defined.

Suppose in a differential equation dy/dx = tan (x + y), the degree is 1, whereas for a differential equation tan (dy/dx) = x + y, the degree is not defined. These type of differential equations can be observed with other trigonometry functions such as sine, cosine and so on.

Let us see some more examples on finding the degree and order of differential equations.

Example 3:-

\(\begin{array}{l}\frac{d^2 y}{dx^2} + cos\frac{d^2 y}{dx^2} = 5x\end{array} \)

The given differential equation is not a polynomial equation in derivatives. Hence, the degree of this equation is not defined.

Example 4:-

\(\begin{array}{l}(\frac{d^3 y}{dx^3})^2 + y = 0\end{array} \)

The order of this equation is 3 and the degree is 2.

Example 5:- Figure out the order and degree of differential equation that can be formed from the equation

\(\begin{array}{l}\sqrt{1 – x^2} + \sqrt{1 – y^2} = k(x – y)\end{array} \)
.

Solution:-

Let

\(\begin{array}{l}x = sin \theta, y = sin \phi \end{array} \)

So, the given equation can be rewritten as

\(\begin{array}{l}\sqrt{1 – sin\theta^2} + \sqrt{1 – sin\phi^2} = k(sin \theta – sin \phi)\end{array} \)

\(\begin{array}{l} \Rightarrow (cos \theta + cos \phi) = k(sin \theta – sin \phi)\end{array} \)

\(\begin{array}{l} \Rightarrow 2 cos \frac{\theta + \phi}{2} cos\frac{\theta – \phi}{2} = 2 k cos \frac{\theta + \phi}{2} sin \frac{\theta – \phi}{2} \end{array} \)

\(\begin{array}{l}cot \frac{\theta – \phi}{2} = k\end{array} \)

\(\begin{array}{l}\theta – \phi = 2cot^{-1} k\end{array} \)

\(\begin{array}{l} sin^{-1}x – sin^{-1}y = 2cot^{-1} k\end{array} \)

Differentiating both sides w. r. t. x, we get

\(\begin{array}{l}\frac{1}{\sqrt{1 – x^2}} – \frac{1}{\sqrt{1 – y^2}}\, \frac{dy}{dx} = 0\end{array} \)

So, the degree of the differential equation is 1 and it is a first order differential equation.

Note: If the DE in which differential coefficient is present inside the parenthesis of any another function as a composite, then first attempt to make it as simple as possible. Now, check whether it in the form of a polynomial in terms of derivatives. If it is a polynomial, the degree can be defined.

In the upcoming discussions, we will learn about solutions to the various forms of differential equations. We here at BYJU’S will help you tackle all your doubts in the easiest possible way. Visit us to enjoy the beauty of simplicity in solving all your doubts.

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