Question

Obtain the differential equation of the family of curves \(y = \frac{2ce^{2x}}{1+ce^{2x}}\).

Hint:

When forming a differential equation, the primary goal is to eliminate the arbitrary constant. It's often easier to first algebraically manipulate the equation to isolate the constant term before differentiating. This can simplify the substitution step significantly.

Answer 1

Step 1: Understanding the Concept:
To find the differential equation for a family of curves, we need to eliminate the arbitrary constant (in this case, 'c') from the given equation. This is typically done by differentiating the equation and then using the original and differentiated equations to eliminate the constant.
Step 2: Key Formula or Approach:
The strategy is to first rearrange the given equation to express the constant 'c' (or a term containing 'c') in terms of x and y. Then, differentiate the original equation with respect to x and substitute the expression for the constant to get an equation involving only x, y, and derivatives of y.
Step 3: Detailed Explanation:
The given equation is: \[ y = \frac{2ce^{2x}}{1+ce^{2x}} \] First, let's try to isolate the term containing the constant 'c'. \[ y(1+ce^{2x}) = 2ce^{2x} \] \[ y + yce^{2x} = 2ce^{2x} \] \[ y = 2ce^{2x} - yce^{2x} \] \[ y = ce^{2x}(2-y) \] From this, we can express the term \(ce^{2x}\) as: \[ ce^{2x} = \frac{y}{2-y} \quad \text{---(1)} \] Now, let's differentiate the original equation \(y = \frac{2ce^{2x}}{1+ce^{2x}}\) with respect to x using the quotient rule: \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}\).
Let \(u = 2ce^{2x}\) and \(v = 1+ce^{2x}\).
Then \(u' = 4ce^{2x}\) and \(v' = 2ce^{2x}\).
\[ \frac{dy}{dx} = \frac{(1+ce^{2x})(4ce^{2x}) - (2ce^{2x})(2ce^{2x})}{(1+ce^{2x})^2} \] \[ \frac{dy}{dx} = \frac{4ce^{2x} + 4(ce^{2x})^2 - 4(ce^{2x})^2}{(1+ce^{2x})^2} \] \[ \frac{dy}{dx} = \frac{4ce^{2x}}{(1+ce^{2x})^2} \quad \text{---(2)} \] Now, substitute the expression for \(ce^{2x}\) from equation (1) into equation (2): \[ \frac{dy}{dx} = \frac{4\left(\frac{y}{2-y}\right)}{\left(1+\frac{y}{2-y}\right)^2} \] Simplify the denominator: \[ 1+\frac{y}{2-y} = \frac{(2-y)+y}{2-y} = \frac{2}{2-y} \] Substitute this back into the equation for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{\frac{4y}{2-y}}{\left(\frac{2}{2-y}\right)^2} = \frac{\frac{4y}{2-y}}{\frac{4}{(2-y)^2}} \] \[ \frac{dy}{dx} = \frac{4y}{2-y} \times \frac{(2-y)^2}{4} \] \[ \frac{dy}{dx} = y(2-y) \] Step 4: Final Answer:
The required differential equation is \(\frac{dy}{dx} = 2y - y^2\).