Solving the Differential Equation \(\ln\left(\frac{dy}{dx}\right)+y = x\)

The given problem asks us to find the solution to a specific differential equation. A differential equation is an equation that relates a function with its derivatives.

The given differential equation is:

\(\ln\left(\frac{dy}{dx}\right)+y = x\)

Our goal is to find the function \(y\) in terms of \(x\) that satisfies this equation. We can start by isolating the derivative term, \(\frac{dy}{dx}\).

Subtract \(y\) from both sides:

\(\ln\left(\frac{dy}{dx}\right) = x - y\)

To eliminate the natural logarithm (\(\ln\)), we can exponentiate both sides with base \(e\). Recall that \(e^{\ln(a)} = a\).

\(e^{\ln\left(\frac{dy}{dx}\right)} = e^{x - y}\)

\(\frac{dy}{dx} = e^{x - y}\)

Using the property of exponents \(e^{a-b} = e^a \cdot e^{-b}\), we can rewrite the right side:

\(\frac{dy}{dx} = e^x \cdot e^{-y}\)

This differential equation is a separable equation because we can separate the variables \(y\) and \(x\) on different sides of the equation. We can multiply both sides by \(e^y\) and multiply both sides by \(dx\).

\(e^y \, dy = e^x \, dx\)

Now, we can integrate both sides of the equation with respect to their respective variables:

\(\int e^y \, dy = \int e^x \, dx\)

The integral of \(e^z\) with respect to \(z\) is \(e^z\) plus a constant of integration. Integrating both sides gives:

\(e^y + C_1 = e^x + C_2\)

Here, \(C_1\) and \(C_2\) are constants of integration. We can combine the constants into a single constant \(C = C_2 - C_1\).

\(e^y = e^x + C\)

To match the format of the given options, we can rearrange the terms:

\(e^x - e^y = -C\)

Since \(C\) is an arbitrary constant, \(-C\) is also an arbitrary constant. Let's call this new constant \(c\).

\(e^x - e^y = c\)

This is the general solution to the given differential equation.

Let's compare this solution with the provided options:

• Option 1: \(e^x + e^y = c\) - Does not match.
• Option 2: \(e^{x + y} = c\) - Does not match (This would be \(e^x e^y = c\)).
• Option 3: \(e^x - e^y = c\) - Matches our derived solution.
• Option 4: \(e^{x - y} = c\) - Does not match (This would be \(e^x / e^y = c\)).

Therefore, the correct solution is \(e^x - e^y = c\).

Revision Table: Key Steps in Solving Separable DEs

Additional Information on Differential Equations and Solutions

A differential equation is an equation containing an unknown function and one or more of its derivatives. These equations are fundamental in describing processes involving change, such as population growth, radioactive decay, and the motion of objects.

• Order: The order of a differential equation is the order of the highest derivative appearing in the equation. The given equation \(\ln\left(\frac{dy}{dx}\right)+y = x\) involves only the first derivative \(\frac{dy}{dx}\), so it is a first-order differential equation.
• Linear vs. Non-linear: A differential equation is linear if the dependent variable and its derivatives appear only in the first degree and are not multiplied together. The given equation involves \(\ln\left(\frac{dy}{dx}\right)\) and \(e^{-y}\) after separation, which makes it a non-linear differential equation.
• Separable Equations: A first-order differential equation of the form \(\frac{dy}{dx} = f(x, y)\) is called separable if \(f(x, y)\) can be written as a product of a function of \(x\) and a function of \(y\), i.e., \(\frac{dy}{dx} = g(x)h(y)\). This allows us to separate variables as \(\frac{dy}{h(y)} = g(x) \, dx\) and integrate both sides. The equation \(\frac{dy}{dx} = e^x e^{-y}\) is a separable equation.
• General Solution: The general solution of a differential equation is a family of functions that satisfies the equation. It typically involves one or more arbitrary constants (like \(c\) in our solution), the number of constants usually equals the order of the differential equation.
• Particular Solution: A particular solution is obtained from the general solution by assigning specific values to the constants, usually determined by initial or boundary conditions.