The order and degree of the differential equation \(\frac{{{d^3}y}}{{d{x^3}}} + 4\sqrt{\left[{{{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}}}\right]}= 0\;\)
3 and 2
Differential Equation Order and Degree
Understanding the order and degree of a differential equation is a fundamental concept in mathematics. Let's break down the given differential equation to determine its order and degree.
The given differential equation is:
\(\frac{{{d^3}y}}{{d{x^3}}} + 4\sqrt{\left[{{{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}}}\right]}= 0\;\)
Order of the Differential Equation
The order of a differential equation is defined as the order of the highest derivative present in the equation. To find the order, we look at all the derivatives and identify the one with the highest order.
- The first term is \(\frac{{{d^3}y}}{{d{x^3}}}\), which is a derivative of order 3.
- The term inside the square root, \(\frac{{dy}}{{dx}}\), is a derivative of order 1.
Comparing the orders, the highest derivative present in the equation is \(\frac{{{d^3}y}}{{d{x^3}}}\), which has an order of 3.
Therefore, the order of the given differential equation is 3.
Degree of the Differential Equation
The degree of a differential equation is defined as the power of the highest order derivative, after the equation has been made free of radicals and fractions involving derivatives. If the equation cannot be expressed as a polynomial in its derivatives, the degree is not defined.
In our given differential equation, there is a square root involving derivatives, which means it is not in polynomial form with respect to its derivatives. We must eliminate this radical to determine the degree.
Let's rearrange the equation to remove the radical:
1. Isolate the radical term:
\[ \frac{{{d^3}y}}{{d{x^3}}} = -4\sqrt{\left[{{{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}}}\right]} \]
2. Square both sides of the equation to eliminate the square root:
\[ \left(\frac{{{d^3}y}}{{d{x^3}}}\right)^2 = \left(-4\sqrt{\left[{{{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}}}\right]}\right)^2 \]
\[ \left(\frac{{{d^3}y}}{{d{x^3}}}\right)^2 = 16\left[{{{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}}}\right] \]
3. Expand the right side:
\[ \left(\frac{{{d^3}y}}{{d{x^3}}}\right)^2 = 16{\left( {\frac{{dy}}{{dx}}} \right)^3} + 16{y^2} \]
Now, the differential equation is expressed as a polynomial in its derivatives. We can clearly identify the powers of the derivatives.
- The highest order derivative is \(\frac{{{d^3}y}}{{d{x^3}}}\).
- The power of this highest order derivative \(\left(\frac{{{d^3}y}}{{d{x^3}}}\right)\) in the polynomial form is 2.
Therefore, the degree of the differential equation is 2.
Summary of Order and Degree
Based on our analysis:
- The order of the differential equation is 3.
- The degree of the differential equation is 2.
Thus, the order and degree of the given differential equation are 3 and 2, respectively.
| Concept | Definition | Value for Given Equation |
|---|---|---|
| Order | The order of the highest derivative present in the equation. | 3 (from \(\frac{{{d^3}y}}{{d{x^3}}}\)) |
| Degree | The power of the highest order derivative after making the equation polynomial in its derivatives. | 2 (from \(\left(\frac{{{d^3}y}}{{d{x^3}}}\right)^2\)) |
Important Questions from Order and Degree of a Differential Equation
Consider the following in respect of the differential equation:
\(\frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\)
1. The degree of the differential equation is 1.
2. The order of the differential equation is 2.
Which of the above statements is/are correct?
The partial differential equation \(\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is a
The degree of the differential equation \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\frac{{dy}}{{dx}}} \right)^2} + \sin x\left( {\frac{{dy}}{{dx}}} \right) + y = 0\) is:
In the following partial differential equation, θ is a function of t and z, and D and K are functions of θ
\(D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} + \frac{{\delta K\left( \theta \right)}}{{\delta z}} - \frac{{\delta \theta }}{{\delta t}} = 0\)
The above equation isThe solution of the equation \({\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0{\rm{}}\) passing through the point (1,1) is
