Question Statement

Find the order and degree, if defined, for the differential equation: $dy-\sin xdx=0$

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Background and Explanation

The order of a differential equation is the order of the highest derivative present in the equation. The degree is the power of the highest order derivative, provided the differential equation is a polynomial in its derivatives.

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Solution

To find the order and degree, we first need to rewrite the given differential equation in a standard form involving derivatives.

Given: $dy-\sin xdx=0$

Step 1: Rearrange the equation Move the term involving $dx$ to the right side: $dy=\sin xdx$

Step 2: Express as a derivative Divide both sides by $dx$ to obtain the derivative $\frac{dy}{dx}$: $\frac{dy}{dx}=\sin x$

Step 3: Determine the Order The highest order derivative present in this equation is $\frac{dy}{dx}$, which is a first-order derivative. $\Rightarrow \text{Order}=1$

Step 4: Determine the Degree The highest order derivative $\left(\frac{dy}{dx}\right)$ is raised to the power of $1$. Since the equation is a polynomial in terms of its derivative, the degree is defined. $\Rightarrow \text{Degree}=1$

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Key Formulas or Methods Used

• Order: The highest derivative $\frac{d^{n}y}{d{x}^n}$ appearing in the equation.
• Degree: The power of the highest order derivative after the equation is cleared of fractions and radicals (relative to the derivatives).

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Summary of Steps

1. Rearrange the differential form $dy-\sin xdx=0$ into the derivative form $\frac{dy}{dx}=\sin x$.
2. Identify the highest derivative present to find the order (1).
3. Identify the exponent of that highest derivative to find the degree (1).