Question

\(e^{(d y / d x)}=x+1\) given that when \(x=0, y=3\)

Short answer

Question: Determine the explicit solution for the first-order differential equation \(e^{(dy/dx)} = x + 1\) with the initial condition that when \(x = 0\), \(y = 3\). Answer: \(y = \ln{\left(\frac{1}{2}x^2 + x + e^3\right)}\)

Step-by-step solution

Rewrite the differential equation

The given equation is \(e^{(d y / d x)}=x+1\) and can be rewritten as a separable equation. To do this, rewrite \(e^{(d y / d x)}\) as \(dy=e^{-y} dx\).

Separate the variables

Our goal is to have \(y\) and \(dy\) on one side of the equation and \(x\) and \(dx\) on the other side. Divide both sides by \(e^{-y}\) and multiply by \(dx\), we get: $$ e^y dy= (x+1)dx $$

Integrate both sides

Now, integrate both sides of the equation with respect to \(x\) and \(y\): $$ \int e^y dy = \int (x+1) dx $$

Find the antiderivatives

Determine the antiderivative for both sides of the equation: $$ e^y = \frac{1}{2}x^2 + x + C $$ where \(C\) is the constant of integration.

Apply initial condition

We are given the initial condition that when \(x = 0\), \(y = 3\). Plug these values into the equation to find the constant \(C\): $$ e^3 = \frac{1}{2}(0)^2 + 0 + C \\ C = e^3 $$

Write the final solution

Substitute the value of \(C\) back into the equation to find the general solution: $$ e^y = \frac{1}{2}x^2 + x + e^3 $$ To find the explicit solution, take the natural logarithm of both sides: $$ y = \ln{\left(\frac{1}{2}x^2 + x + e^3\right)} $$ Note that this is the explicit solution for the given differential equation subject to the initial condition (\(x = 0\), \(y = 3\)).

Key concepts

Separable Differential Equations

Separable differential equations are a class of ordinary differential equations (ODEs) that can be separated into two sides involving different variables, each side with one variable and its differential only. The process can be visualized as untangling the equation to isolate each variable on one side, enabling us to integrate each part with ease.

Considering the example given, the ODE \(e^{(dy/dx)} = x + 1\) initially doesn't seem to be separable. However, the first step taken in the solution involves manipulation to achieve that form: \(dy = e^{-y} dx\), which sets the stage for separation with \(y\) and \(dy\) on one side and \(x\) and \(dx\) on the other. This manipulation brings the equation to a point where integration is possible.

Integrating Factors

An integrating factor is a function used to transform a non-exact differential equation into an exact one, thus making it easier to solve. It strategically simplifies the equation such that the left and right sides can clearly be identified as derivatives of a product involving the integrating factor.

In this particular exercise, we do not utilize an integrating factor because the equation, after separation, leads directly to integrable forms on both sides. However, it's important to recognize integrating factors as invaluable tools for solving more complex differential equations that cannot be easily separated.

Initial Value Problems

An initial value problem (IVP) is a differential equation accompanied by a specific value, called the initial condition, which the solution must satisfy at a particular point. The inclusion of the initial condition allows us to determine the exact solution that fits the given scenario as opposed to a general family of solutions.

In our exercise, the IVP is presented by the initial condition \(x = 0\), \(y = 3\). Solving the general solution at this point provides the particular constant \(C = e^3\) which, when substituted back into the equation, gives the unique solution that passes through the point \(0, 3\).

Exponential Functions

Exponential functions are the bread and butter of many differential equations and are defined as \( f(x)=a^x\), with \e\ being a commonly used base due to its natural properties. They are characterized by a rate of growth proportional to the value of the function itself.

The original equation in our problem utilizes an exponential function with base \e\, and the process of solving the differential equation inherently involves manipulating and integrating this exponential function. This results in the exponential term \( e^y\), highlighting the interplay between exponents and logarithms in differential equations.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse of differentiation. Finding the antiderivative of a term involves determining the function whose derivative would give the original term. Symbolically, this is represented by the integral sign followed by the function to be integrated.

In our worked-through example, calculating the antiderivatives for \( e^y dy\) and \( (x+1) dx\) directly translates to finding the functions who, when derived, give us the terms we started with. The solution step explicitly demonstrates the evaluation of these antiderivatives and introduces the integration constant \C\, typical in indefinite integration.

Natural Logarithm

The natural logarithm, denoted as \ln x\, is the inverse function of the exponential function \(e^x\). In essence, it helps us 'undo' exponential growth and provides a technique for solving equations where the variable of interest is in the exponent, which is especially common in solving differential equations.

In the final solution of the problem, taking the natural logarithm of both sides of the equation \( e^y = \frac{1}{2}x^2 + x + e^3\) allows us to solve for \(y\). This operation is essential because it provides access to \(y\) in its explicit form, making interpretation and further calculus applications approachable.