Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a

Q: Wait, why do we use the Leibniz Rule here? Can't we just integrate the expression?

Great question! We aren't trying to find the area; we are trying to find the function that *creates* that area. The Leibniz Rule is the inverse operation of integration, allowing us to 'peel back' the integral to reveal the function $f(x)$ inside. Think of it as undoing the accumulation to see the rate of change.

Q: How do I know how many times to differentiate?

Look at the constants! The number of arbitrary constants in your general solution tells you the order of the differential equation. Two constants ($c_1, c_2$) mean you need a second-order differential equation, so two differentiations are mandatory to eliminate them.

Q: The final equation looks different from my result. How do I handle the $e^x$ term?

That's just for algebraic elegance! The options provided don't have negative exponents. Multiplying by $e^x$ is a standard technique to clear out terms like $e^{-x}$ and make the equation look clean and professional. It doesn't change the validity of the equation, just its form.

Question

Let $f(x)$ be a positive function such that the area bounded by $y = f(x), y = 0$ from $x = 0$ to $x = a > 0$ is $e^{-a} + 4a^2 + a - 1$. Then the differential equation, whose general solution is $y = c_1 f(x) + c_2$, where $c_1$ and $c_2$ are arbitrary constants, is

Options:

(A) $(8e^x - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$

(B) $(8e^x - 1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$

(C) $(8e^x + 1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$

(D) $(8e^x + 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a graph, looking at a curve defined by a positive function $f(x)$. You are told that the area under this curve, from $x = 0$ to $x = a$, is given by the expression $e^{-a} + 4a^2 + a - 1$.

This is not just a random collection of terms; it is the accumulated history of the function $f(x)$. To find the function itself, we need to reverse the process of integration.

This is where the Newton-Leibniz Rule becomes our most powerful tool. By differentiating the integral $\int_{0}^{a} f(x) dx$ with respect to the upper limit $a$, we are essentially asking, "How does the area change as the boundary moves?"

The answer, as the Fundamental Theorem of Calculus tells us, is simply the value of the function at that boundary: $f(a)$.

### The Function Revealed

Let us perform this differentiation with precision. We differentiate the expression $e^{-a} + 4a^2 + a - 1$ with respect to $a$.

The derivative of $e^{-a}$ is $-e^{-a}$. The derivative of $4a^2$ is $8a$. The derivative of $a$ is $1$, and the constant $-1$ vanishes into zero.

Thus, we find:

$$f(a) = -e^{-a} + 8a + 1$$

Since $a$ is just a placeholder for our variable, we can confidently state that $f(x) = -e^{-x} + 8x + 1$. We have successfully unmasked the function!

### The General Solution and the Elimination Game

Now, we turn our attention to the second part of our journey. We are given the general solution $y = c_1 f(x) + c_2$. This equation describes a family of curves, and our goal is to find the differential equation that governs them.

Because we have two arbitrary constants, $c_1$ and $c_2$, we know we need a second-order differential equation. We must differentiate $y$ twice.

First, we find the first derivative:

$$\frac{dy}{dx} = c_1 f'(x) = c_1(e^{-x} + 8)$$

Next, we differentiate again to find the second derivative:

$$\frac{d^2y}{dx^2} = c_1(-e^{-x})$$

Now, we face the challenge of eliminating the constant $c_1$. The most elegant way to do this is to take the ratio of the second derivative to the first derivative.

This causes $c_1$ to cancel out entirely, leaving us with a relationship between the derivatives:

$$\frac{d^2y/dx^2}{dy/dx} = \frac{-e^{-x}}{e^{-x} + 8}$$

### The Final Polish

We are almost at the finish line. Cross-multiplying gives us:

$$(e^{-x} + 8) \frac{d^2y}{dx^2} = -e^{-x} \frac{dy}{dx}$$

Rearranging this, we get:

$$(e^{-x} + 8) \frac{d^2y}{dx^2} + e^{-x} \frac{dy}{dx} = 0$$

To match the clean, professional form, we multiply the entire equation by $e^x$. This transforms $e^{-x} \cdot e^x$ into $1$, and $8 \cdot e^x$ into $8e^x$.

The result is the beautiful, final differential equation:

$$(8e^x + 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$$

You have navigated the path from an area integral to a differential equation, mastering the Leibniz Rule and the art of constant elimination along the way. Keep this logical flow in your toolkit, and no differential equation will ever intimidate you again.

Select Answer:

Visualized Solution

Visualizing the Area Under the Curve

Applying the Leibniz Rule

Differentiating the Expression

Defining the Function $f (x)$

The General Solution $y = c_{1} f (x) + c_{2}$

Calculating the First Derivative

Calculating the Second Derivative

Eliminating the Constant $c_{1}$

Simplifying the Expression

Multiplying by $e^{x}$ for Final Form

Final Conclusion

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

The Function Revealed

The General Solution and the Elimination Game

The Final Polish

Similar Questions

The differential equation satisfied by the system of parabolas $y^{2} = 4 a (x + a)$ is:

The differential equation of the family of curves, $x^{2} = 4 b (y + b)$ , $b \in R$ , is

Let $y = f (x)$ be a curve passing through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve lies in the first quadrant and has area 2. Form the differential equation and determine all such possible curves.

The differential equation of the family of curves, $x^{2} = 4 b (y + b)$ , $b \in R$ , is:

The differential equation which represents the family of curves $y = c_{1} e^{c_{2} x}$ , where $c_{1}$ and $c_{2}$ are arbitrary constants, is

The differential equation for the family of circle $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is

The differential equation of all circles passing through the origin and having their centres on the $x$ -axis is

The differential equation whose solution is $A x^{2} + B y^{2} = 1$ where $A$ and $B$ are arbitrary constants is of

The differential equation representing the family of curves $y^{2} = 2 c (x + c)$ , where $c$ is a positive parameter, is of

The differential equation of the family of circles passing through the origin and having centre at the line $y = x$ is :

Select Answer:

Visualized Solution

Visualizing the Area Under the Curve

Applying the Leibniz Rule

Differentiating the Expression

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The General Solution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=c1​f(x)+c2​

Calculating the First Derivative

Calculating the Second Derivative

Eliminating the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c1​

Simplifying the Expression

Multiplying by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ex for Final Form

Final Conclusion

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

The Function Revealed

The General Solution and the Elimination Game

The Final Polish

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation satisfied by the system of parabolas y2=4a(x+a) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation of the family of curves, x2=4b(y+b), b∈R, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation of the family of curves, x2=4b(y+b), b∈R, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation which represents the family of curves y=c1​ec2​x, where c1​ and c2​ are arbitrary constants, is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation of all circles passing through the origin and having their centres on the x-axis is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation representing the family of curves y2=2c(x+c​), where c is a positive parameter, is of

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation of the family of circles passing through the origin and having centre at the line y=x is :