What is the degree of the following differential equation? \(\rm x=\sqrt{1+\frac{d^2y}{dx^2}}\)
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Finding the Degree of a Differential Equation
Let's determine the degree of the given differential equation:
\[ \rm x=\sqrt{1+\frac{d^2y}{dx^2}} \]To find the degree of a differential equation, we first need to understand what order and degree mean.
- Order: The order of a differential equation is the order of the highest derivative present in the equation.
- Degree: The degree of a differential equation is the power of the highest-order derivative, provided that the equation has been made free from radicals and fractions with respect to the derivatives. If the equation cannot be expressed as a polynomial in terms of its derivatives, the degree is undefined.
Step-by-Step Solution
Let's analyze the given equation:
\[ \rm x=\sqrt{1+\frac{d^2y}{dx^2}} \]- Identify the highest-order derivative: The highest derivative present in the equation is \(\frac{d^2y}{dx^2}\).
- Determine the order: Since the highest derivative is \(\frac{d^2y}{dx^2}\), the order of the differential equation is 2.
- Eliminate the radical: The equation contains a square root involving the derivative term. To find the degree, we must remove this radical by squaring both sides of the equation: \[ \rm (x)^2 = \left(\sqrt{1+\frac{d^2y}{dx^2}}\right)^2 \] \[ \rm x^2 = 1+\frac{d^2y}{dx^2} \]
- Rearrange the equation (optional): We can rewrite the equation to make it look like a polynomial in terms of the derivatives: \[ \frac{d^2y}{dx^2} + 1 - x^2 = 0 \] Now, the equation is free from radicals and fractions with respect to the derivatives. It is a polynomial equation in terms of \(\frac{d^2y}{dx^2}\).
- Determine the power of the highest-order derivative: The highest-order derivative is \(\frac{d^2y}{dx^2}\). Its power in the equation \(\frac{d^2y}{dx^2} + 1 - x^2 = 0\) is 1.
Therefore, the degree of the differential equation is 1.
| Concept | Value for \( \rm x=\sqrt{1+\frac{d^2y}{dx^2}} \) |
|---|---|
| Highest Derivative | \( \frac{d^2y}{dx^2} \) |
| Order | 2 |
| Equation without radical | \( \frac{d^2y}{dx^2} = x^2 - 1 \) |
| Power of Highest Derivative | 1 |
| Degree | 1 |
Comparing this result with the given options, we find that the degree is 1.
Revision Table: Order vs. Degree
| Feature | Order of a Differential Equation | Degree of a Differential Equation |
|---|---|---|
| Definition | Order of the highest derivative. | Power of the highest-order derivative after removing radicals/fractions w.r.t. derivatives. |
| Requirement for Definition | Always defined if derivatives exist. | Defined only if the equation is a polynomial in derivatives. |
| How to Find | Look for the highest \( n \) in \( \frac{d^ny}{dx^n} \). | Clear radicals and fractions involving derivatives, then find the exponent of the highest derivative term. |
Additional Information: Types of Differential Equations
Differential equations are classified based on various characteristics. Understanding these classifications helps in solving them.
- Based on Type:
- Ordinary Differential Equation (ODE): Contains only ordinary derivatives with respect to a single independent variable (like \( \frac{dy}{dx} \)). The given equation is an ODE.
- Partial Differential Equation (PDE): Contains partial derivatives with respect to two or more independent variables (like \( \frac{\partial z}{\partial x} \)).
- Based on Order: Classified by the order of the highest derivative (e.g., first-order, second-order). The given equation is a second-order ODE.
- Based on Linearity:
- Linear Differential Equation: Can be written in the form \( a_n(x) \frac{d^ny}{dx^n} + a_{n-1}(x) \frac{d^{n-1}y}{dx^{n-1}} + ... + a_1(x) \frac{dy}{dx} + a_0(x) y = b(x) \). The dependent variable \( y \) and its derivatives appear only in the first degree and are not multiplied together.
- Non-linear Differential Equation: Any differential equation that is not linear. The given equation \( \frac{d^2y}{dx^2} = x^2 - 1 \) is linear because \( y \) and its derivatives appear only to the power of 1 and are not multiplied.
Similar Questions
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Consider the following in respect of the differential equation:
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