The differential equation whose solution is Ax² + By² = 1 where A and B are arbitrary constants is o

Q: Wait, why is the order exactly 2? Can't we just differentiate once and get a first-order equation?

That is a fantastic question! The order of a differential equation is defined by the number of independent arbitrary constants in the general solution. Because your equation $Ax^2 + By^2 = 1$ has two constants, $A$ and $B$, you have two degrees of freedom. If you only differentiate once, you will still have one constant left over, meaning you haven't fully 'formed' the differential equation for the entire family. You must differentiate twice to eliminate both constants completely.

Q: How do we determine the degree? Is it just the highest power in the equation?

Not quite! The degree is defined as the power of the *highest-order derivative* present in the differential equation, *after* the equation has been written as a polynomial in terms of derivatives. In our final equation, $xyy'' + x(y')^2 - yy' = 0$, the highest derivative is $y''$. The power of this $y''$ is 1. Therefore, the degree is 1.

Q: Why did we divide by $B$ in the final step? Is that always allowed?

Yes, in this context, it is! Since $A$ and $B$ are arbitrary constants defining a family of curves, we assume $B \neq 0$ (otherwise, the equation would collapse into something else, like $Ax^2 = 1$, which is not the general family we are analyzing). Dividing by $B$ is a standard algebraic move to simplify the expression and isolate the relationship between $x$, $y$, and the derivatives.

Question

The differential equation whose solution is $Ax^2 + By^2 = 1$ where $A$ and $B$ are arbitrary constants is of

Options:

(A) second order and second degree

(B) first order and second degree

(C) first order and first degree

(D) second order and first degree

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a vast canvas, and on this canvas, you are drawing a family of ellipses defined by the equation:

$$Ax^2 + By^2 = 1$$

These are not just static shapes; they are a dynamic family, shifting and stretching as you vary the arbitrary constants $A$ and $B$. In the realm of calculus, we seek the 'DNA' of this family—the differential equation that governs every single one of them.

### The Golden Rule of Order

Before we dive into the algebra, let's appreciate a fundamental truth in differential equations. The order of a differential equation is a direct reflection of the number of independent arbitrary constants in the general solution.

Think of it as a conservation law: if you have two constants, $A$ and $B$, you have two 'degrees of freedom' that must be constrained. To eliminate two constants, you must differentiate twice.

We know, before we even pick up our pen, that our final result must be a **second-order differential equation**.

### The Calculus Journey: Differentiating Twice

Our starting point is the equation $Ax^2 + By^2 = 1$. To find the relationship between the variables and their first derivative, we differentiate with respect to $x$:

$$\frac{d}{dx}(Ax^2) + \frac{d}{dx}(By^2) = \frac{d}{dx}(1)$$

This yields $2Ax + 2Byy' = 0$. Dividing by 2, we arrive at our first milestone:

$$Ax + Byy' = 0$$

Here, $y'$ is our shorthand for $\frac{dy}{dx}$. Now, we must differentiate again with respect to $x$. The derivative of $Ax$ is $A$, and for the term $Byy'$, we apply the product rule:

$$A + B(y y'' + (y')^2) = 0$$

### The Art of Elimination

We now have two equations:
(1) $Ax + Byy' = 0$
(2) $A + B(y y'' + (y')^2) = 0$

From the first equation, we isolate $A$:

$$A = -\frac{Byy'}{x}$$

Substituting this into the second equation yields:

$$-\frac{Byy'}{x} + B(y y'' + (y')^2) = 0$$

Since $B$ is a non-zero constant, we divide the entire equation by $B$ to eliminate it:

$$-\frac{yy'}{x} + y y'' + (y')^2 = 0$$

To remove the fraction, we multiply the entire equation by $x$:

$$-yy' + xyy'' + x(y')^2 = 0$$

### The Final Verdict

Rearranging the terms, we obtain the final, pristine differential equation:

$$xyy'' + x(y')^2 - yy' = 0$$

The highest derivative present is $y''$, which confirms our initial prediction: the **order is 2**. Since the power of this highest derivative is 1, the **degree is 1**.

You have successfully distilled the family of curves into its governing differential equation, mastering the geometry, calculus, and algebra required for this classic problem.

Animated Solution for Mathematics - Differential Equations: The differential equation whose solution is $A x^{2} + B y^{2} = 1$ where $A$ and $B$ are arbitrary constants is of

Select Answer:

Visualized Solution

Identify the Family of Curves

Determine the Order

First Differentiation

Second Differentiation

Eliminate Constant A

Final Simplification

Conclusion: Order and Degree

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

The Golden Rule of Order

The Calculus Journey

The Art of Elimination

The Final Verdict

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation whose solution is Ax2+By2=1 where A and B are arbitrary constants is of