Question

The differential equation which represents the family of curves \(\mathrm{y}=\mathrm{c}_{1} \mathrm{e}^{(\mathrm{c}) 2 \mathrm{x}}\), where \(\mathrm{c}_{1}\) and \(\mathrm{c}_{2}\) are arbitarary constants, is: (A) \(y^{\prime}=y^{2}\) (B) \(\mathrm{y}^{\prime \prime}=\mathrm{y}^{\prime} \mathrm{y}\) (C) \(\mathrm{yy}^{\prime \prime}=\left(\mathrm{y}^{1}\right)^{2}\) (D) \(\mathrm{yy}^{\prime \prime}=\mathrm{y}^{\prime}\)

Short answer

The short answer is: (C) \(yy''= (y^1)^2\).

Step-by-step solution

Differentiate with respect to x.

Differentiate the given equation y = c_1 e^{c_2 2x} with respect to x to find the first derivative: \(y' = \frac{dy}{dx} = c_1 (2c_2) e^{c_2 2x}\)

Differentiate again with respect to x.

Differentiate \(y'\) with respect to x to find the second derivative: \(y'' = \frac{d^2y}{dx^2} = c_1 (2c_2)^2 e^{c_2 2x}\)

Eliminate arbitrary constants c_1 and c_2.

We have: \(y = c_1 e^{c_2 2x}\) \(y' = c_1 (2c_2) e^{c_2 2x}\) \(y'' = c_1 (2c_2)^2 e^{c_2 2x}\) To eliminate c_1 and c_2, we can use the second equation and divide it by the first equation and then divide the third equation by the second equation. \(\frac{y'}{y} = 2c_2\) \(\frac{y''}{y'} = 2c_2\) Since \(\frac{y'}{y} = \frac{y''}{y'}\), we can rewrite the equation as: \(y' y = y''y\) Now, we check our options: (A) \(y' = y^2\) This is not a match because there is no \(y^2\) term in our transformed equation. (B) \(y''= y' y\) This is not a match because there is a product term missing from our transformed equation. (C) \(y y''= (y^1)^2\) This is a match with our transformed equation \(y' y = y''y\)! (D) \(yy''= y'\) This is not a match because our transformed equation has equal signs on both sides.

Conclusion

Based on our transformed differential equation and the given options, we can conclude that the differential equation which represents the family of curves \(y = c_1 e^{c_2 2x}\) is: (C) \(y y''= (y^1)^2\)

Key concepts

First Derivative

To begin understanding the role of the first derivative in solving differential equations, let's break it down. When dealing with a function, the first derivative represents the rate at which the function's output changes with respect to its input. Simply put, it tells us how fast the function values are increasing or decreasing.

For the given problem, we have the expression \(y = c_1 e^{c_2 2x}\). When we differentiate this expression with respect to \(x\), we obtain the first derivative, \(y'\), which describes how the exponential function is changing as \(x\) varies.

• First, focus on the constant \(c_2\) within the exponent. Differentiating the exponential \(e^{c_2 2x}\) involves multiplying by the derivative of \(c_2 2x\), which is simply \(2c_2\).
• Thus, differentiating gives us \(y' = c_1 (2c_2) e^{c_2 2x}\).

The first derivative helps set the stage for further differentiations, moving us closer to solving the differential equation.

Second Derivative

Continuing with the differentiation, the second derivative takes analysis one step further. By finding the second derivative of a function, you measure how the rate of change itself is changing. In physical terms, if the first derivative is like velocity, then the second derivative is analogous to acceleration.

When we differentiate the first derivative \(y'\) again with respect to \(x\), we arrive at the second derivative \(y''\). For our function, this involves:

• Starting from \(y' = c_1 (2c_2) e^{c_2 2x}\), and differentiating it gives us \(y'' = c_1 (2c_2)^2 e^{c_2 2x}\).
• The constants \(2c_2\) multiply again, reflecting the squared term in \(y''\).

The second derivative thus provides us with another perspective on the behaviour of the exponential curve defined by our original function. It plays a crucial role in constructing the differential equation without explicit arbitrary constants.

Eliminate Constants

The final step in solving the differential equation involves eliminating arbitrary constants, in this case, \(c_1\) and \(c_2\). This step turns the parametric description of the curve family into a more universal equation.

To eliminate these constants, observe the relationships between \(y\), \(y'\), and \(y''\):

• By dividing \(y' = c_1 (2c_2) e^{c_2 2x}\) by \(y = c_1 e^{c_2 2x}\), we simplify to \(\frac{y'}{y} = 2c_2\).
• Next, involve the second derivative; dividing \(y'' = c_1 (2c_2)^2 e^{c_2 2x}\) by \(y'\) gives \(\frac{y''}{y'} = 2c_2\).

Because both of these expressions equal \(2c_2\), we equate them to form \(\frac{y'}{y} = \frac{y''}{y'}\), leading to a simplified equation that allows for the elimination of constants: \(y' y = y'' y\).

This step effectively purifies the equation into its simplest form, revealing the specific differential equation that characterizes the family of curves without relying on the constants \(c_1\) and \(c_2\). This conversion is essential for matching with the standard form of differential equations. The solution was found to be option (C): \(yy'' = (y^1)^2\).