The degree of the differential equation is :
2
Finding the Degree of a Differential Equation
To determine the degree of a differential equation, we first need to ensure the equation is expressed as a polynomial in the derivatives. This involves clearing any radicals or fractional powers involving the derivatives.
The given differential equation is:
\[\begin{equation*} \left[ 1- \frac{dy}{dx}\right]^{3/2} = k\frac{d^2y}{dx^2} \end{equation*}\]This equation involves a fractional power (3/2) on the term containing the first derivative, $\frac{dy}{dx}$. To eliminate this fractional power, we need to raise both sides of the equation to the power of 2.
Squaring both sides of the equation:
\[\begin{equation*} \left( \left[ 1- \frac{dy}{dx}\right]^{3/2} \right)^2 = \left( k\frac{d^2y}{dx^2} \right)^2 \end{equation*}\]This simplifies to:
\[\begin{equation*} \left( 1- \frac{dy}{dx}\right)^{3} = k^2\left(\frac{d^2y}{dx^2}\right)^2 \end{equation*}\]Now the equation is free from fractional powers involving derivatives. Let's examine the terms involving derivatives:
- The first derivative is $\frac{dy}{dx}$. This appears inside the term $(1 - \frac{dy}{dx})^3$. If we expand this, the highest power of $\frac{dy}{dx}$ would be 3.
- The second derivative is $\frac{d^2y}{dx^2}$. This appears as $(\frac{d^2y}{dx^2})^2$. The power of this term is 2.
The order of a differential equation is the order of the highest derivative present in the equation. In this equation, the highest order derivative is $\frac{d^2y}{dx^2}$, which is a second-order derivative. Therefore, the order of this differential equation is 2.
The degree of a differential equation is the power of the highest order derivative after the equation has been made free from radicals and fractions involving derivatives. In the equation we obtained after squaring:
\[\begin{equation*} \left( 1- \frac{dy}{dx}\right)^{3} = k^2\left(\frac{d^2y}{dx^2}\right)^2 \end{equation*}\]The highest order derivative is $\frac{d^2y}{dx^2}$. Its power in this equation is 2.
Thus, the degree of the given differential equation is 2.
Understanding Order and Degree
It's important to distinguish between the order and degree of a differential equation:
- Order: The highest order of the derivative appearing in the differential equation.
- Degree: The power of the highest order derivative after the equation is cleared of fractions and radicals involving derivatives. The equation must be expressible as a polynomial in the derivatives for the degree to be defined.
Step-by-Step Solution
Here are the steps followed to find the degree:
- Identify the derivatives present: $\frac{dy}{dx}$ (order 1) and $\frac{d^2y}{dx^2}$ (order 2).
- Identify any terms with fractional or radical powers involving derivatives: The term $[1 - \frac{dy}{dx}]^{3/2}$ has a fractional power.
- Eliminate fractional powers by raising both sides to an appropriate power: Square both sides to remove the 3/2 power. The equation becomes $\left( 1- \frac{dy}{dx}\right)^{3} = k^2\left(\frac{d^2y}{dx^2}\right)^2$.
- Identify the highest order derivative in the resulting equation: This is $\frac{d^2y}{dx^2}$.
- Determine the power of this highest order derivative: The term is $(\frac{d^2y}{dx^2})^2$. The power is 2.
- This power is the degree of the differential equation.
| Concept | Definition |
|---|---|
| Order of a Differential Equation | The order of the highest derivative in the equation. |
| Degree of a Differential Equation | The power of the highest order derivative, after clearing fractions and radicals involving derivatives. |
Revision Table: Key Concepts
| Term | Meaning | Example (from this problem) |
|---|---|---|
| Differential Equation | An equation involving independent variable(s), dependent variable(s), and their derivatives. | $[ 1- {dy/dx}]^{3/2} = kd^2y/dx^2$ |
| Derivative | Rate of change of a function. Examples: $dy/dx$, $d^2y/dx^2$. | $dy/dx$, $d^2y/dx^2$ |
| Order | Highest order of differentiation. | 2 (from $d^2y/dx^2$) |
| Degree | Power of the highest order derivative (after clearing radicals/fractions). | 2 (from $(d^2y/dx^2)^2$ after squaring) |
Additional Information: Polynomial Form
The definition of degree requires the differential equation to be expressible as a polynomial in the derivatives. This means that after clearing fractions and radicals, the equation should look like a polynomial where the variables are the derivatives ($\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, etc.). For example, an equation like $e^{dy/dx} + y = 0$ does not have a defined degree because $e^{dy/dx}$ cannot be expressed as a polynomial in $dy/dx$. Our equation, after squaring, is $\left( 1- \frac{dy}{dx}\right)^{3} - k^2\left(\frac{d^2y}{dx^2}\right)^2 = 0$, which is a polynomial in $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.
The degree is the highest power among the terms involving the highest order derivative. In our case, the highest order derivative is $\frac{d^2y}{dx^2}$, and its highest power is 2.
2
The equation contains a fractional power, so we first remove it by squaring both sides:
\[ \left(1 - \frac{dy}{dx} \right)^3 = k^2 \left( \frac{d^2y}{dx^2} \right)^2 \]
Now the equation is free from radicals and fractions involving derivatives.
Highest order derivative: \( \frac{d^2y}{dx^2} \)
Power of that term = 2
Answer:
2
2
Step 1: Recall Definition of Degree
The degree of a differential equation is the highest power of the highest order derivative present in the equation, after the equation has been made rational and integral in all its derivatives.
Step 2: Rationalize the Equation
Square both sides to eliminate the fractional exponent: \[ \left(\left[1 - \left(\frac{dy}{dx}\right)\right]^{3/2}\right)^2 = \left(k \frac{d^2y}{dx^2}\right)^2 \] \[ \left[1 - \left(\frac{dy}{dx}\right)\right]^3 = k^2 \left(\frac{d^2y}{dx^2}\right)^2 \]
Step 3: Identify the Highest Order Derivative
The highest order derivative is \(\frac{d^2y}{dx^2}\) (second derivative).
Step 4: Determine its Power
The power of \(\frac{d^2y}{dx^2}\) in the rationalized equation is 2.
Step 5: Verify the Equation is Polynomial in Derivatives
The equation is now polynomial in its derivatives: \[ \left(1 - y'\right)^3 = k^2 (y'')^2 \] where \( y' = \frac{dy}{dx} \) and \( y'' = \frac{d^2y}{dx^2} \).
Final Answer:
The degree of the differential equation is \[ \boxed{2} \].
Key Points:
- Original equation had a fractional exponent (3/2)
- Squaring both sides made it rational
- Highest order derivative (\(\frac{d^2y}{dx^2}\)) appears with power 2
- No higher powers of derivatives exist in the equation
Important Questions from Differential Equations
Anubhav spent 14% of his income on electricity bills, 28% on rent and 18% on shopping. If 4/5 of the remaining amount is ₹ 5120, how much did he spend on electricity bills?
Arun's speed of swimming in still water is 5 km/hr. He swims between two points in a river and returns back to the same starting point. He took 20 minutes more to cover the distance upstream than downstream. If the speed of the stream is 2 km/hr, then the distance between the two points is :
If \( f(x) = 2 \left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \), then \( f(x) \) is:
Match List-I with List-II:
List-I List-II (A) Integrating factor of \( xdy - (y + 2x^2)dx = 0 \) (I) \( \frac{1}{x} \) (B) Integrating factor of \( (2x^2 - 3y)dx = xdy \) (II) \( x \) (C) Integrating factor of \( (2y + 3x^2)dx + xdy = 0 \) (III) \( x^2 \) (D) Integrating factor of \( 2xdy + (3x^3 + 2y)dx = 0 \) (IV) \( x^3 \) Choose the correct answer from the options given below:
If t = e2x and y = loge(t2), then d2y/dx2 is :
