The differential equation representing the family of curves y² = 2c(x + √c), where c 0, is a param

Q: Why is the order 1? I see $y^2$ and $y'$ in the equation.

Order is defined by the highest derivative present. Here, the highest derivative is $y'$, which is the first derivative. The power of $y$ or $y'$ does not change the order; only the derivative itself matters.

Q: Why do we have to square the equation? Can't we just look at the equation with the square root?

The definition of the degree of a differential equation requires it to be a polynomial in its derivatives. A square root makes it non-polynomial. Squaring is the algebraic step to reach the standard polynomial form.

Q: Is it possible for the degree to be higher than the order?

Absolutely! Order and degree are independent properties. Order tells you the highest derivative, while degree tells you the highest power of that derivative once the equation is in polynomial form.

Question

The differential equation representing the family of curves $y^2 = 2c(x + \sqrt{c})$, where $c > 0$, is a parameter, is of order and degree as follows :

Options:

(A) order 1, degree 2

(B) order 1, degree 1

(C) order 1, degree 3

(D) order 2, degree 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the differential equation for the family of curves defined by the equation:

$$y^2 = 2c(x + \sqrt{c})$$

Since there is only one parameter, $c$, we know intuitively that the order of our differential equation will be $1$. Our primary goal is to eliminate the parameter $c$ to find the governing rule.

### The Parameter Hunt

We begin by differentiating the equation $y^2 = 2c(x + \sqrt{c})$ with respect to $x$. Applying the chain rule to the left side, we obtain:

$$2yy' = 2c(1 + 0)$$

This simplifies beautifully to:

$$yy' = c$$

This result serves as our golden key, allowing us to express the parameter $c$ directly in terms of $y$ and $y'$.

### The Algebraic Dance

Next, we substitute $c = yy'$ back into the original equation. Replacing every instance of $c$ yields:

$$y^2 = 2(yy')(x + \sqrt{yy'})$$

Assuming $y 
eq 0$, we divide both sides by $y$ to simplify the expression:

$$y = 2y'(x + \sqrt{yy'})$$

We are now close to our goal, but we must address the radical term $\sqrt{yy'}$.

### The Radical Trap

To define the degree of a differential equation, it must be expressed as a polynomial in its derivatives. We cannot leave the equation in its current form with a square root.

First, we isolate the radical term by moving $2xy'$ to the left side:

$$y - 2xy' = 2y'\sqrt{yy'}$$

Now, we square both sides to eliminate the radical:

$$(y - 2xy')^2 = (2y'\sqrt{yy'})^2$$

Expanding the right side, we arrive at:

$$(y - 2xy')^2 = 4(y')^2(yy')$$

This simplifies to the final form:

$$(y - 2xy')^2 = 4y(y')^3$$

### Final Conclusion

We have successfully derived a clean, polynomial differential equation:

$$(y - 2xy')^2 = 4y(y')^3$$

The highest derivative present is $y'$, confirming the **order is 1**. The highest power of the derivative $y'$ is $3$, confirming the **degree is 3**.

Animated Solution for Mathematics - Differential Equations: The differential equation representing the family of curves $y^{2} = 2 c (x + c)$ , where $c > 0$ , is a parameter, is of order and degree as follows :

Select Answer:

Visualized Solution

Family of Curves $y^{2} = 2 c (x + c)$

Eliminating the Parameter $c$

Differentiating w.r.t. $x$

Expressing $c$ in terms of $y^{'}$

Substituting $c$ back

Simplifying the Equation

Isolating the Radical Term

Squaring to Remove Radicals

Identifying Order and Degree

The Sigma Insight: Order and Degree of a Differential Equation

Analyzing the Setup

The Parameter Hunt

The Algebraic Dance

The Radical Trap

Final Conclusion

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