Question

Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$\frac{d y}{d x}=e^{x-y}, y(0)=\ln 3$$

Short answer

Answer: The solution to the given initial value problem is \(y(x) = \ln \left(-\frac{1}{e^x - \frac{2}{3}}\right)\).

Step-by-step solution

Determine if the equation is separable

To determine whether the given differential equation is separable, we must rewrite it as a product of two functions, one of \(x\) and one of \(y\). Let's try to rewrite the given equation as: $$\frac{dy}{dx} = e^{x-y}$$ We want to rewrite the equation in the form: $$\frac{dy}{g(y)} = h(x) dx$$ We can achieve this by writing: $$\frac{dy}{e^y} = e^x dx$$ The equation is separable since it is in the form \(\frac{1}{g(y)}\frac{dy}{dx}=h(x)\).

Integrate both sides

Now that we have separated the variables, let's integrate both sides of the equation with respect to their corresponding variables: $$\int \frac{dy}{e^y} = \int e^x dx$$ By applying integration rules, we obtain: $$-\int e^{-y} dy = \int e^x dx$$ Now integrate: $$-e^{-y} = e^x + C_1$$

Solve for y

Now, we will isolate y by following these steps: 1. Multiply both sides by -1: $$e^{-y} = -e^x - C_1$$ 2. Take the reciprocal of both sides: $$e^y = -\frac{1}{e^x + C_1}$$ 3. Take the natural logarithm of both sides to get y: $$y = \ln \left(-\frac{1}{e^x + C_1}\right)$$

Apply initial condition to find the constant of integration

Now we will apply the initial condition \(y(0)=\ln 3\) to find the constant of integration \(C_1\): $$\ln 3 = \ln \left(-\frac{1}{e^0 + C_1}\right)$$ Taking the exponential of both sides to get rid of the natural logarithm: $$3 = -\frac{1}{1 + C_1}$$ Solve for \(C_1\): $$C_1 = -\frac{2}{3}$$ Now we can substitute this value back into the general solution: $$y(x) = \ln \left(-\frac{1}{e^x - \frac{2}{3}}\right)$$ This is the solution to the given initial value problem.

Key concepts

Separable Differential Equations

Separable differential equations are a special class of ordinary differential equations where the variables can be separated on different sides of the equation. This separation transforms the problem into one where we can perform integration on each variable independently.

Consider the equation \( \frac{dy}{dx} = e^{x-y} \). It is separable because we can isolate the variables as follows: \( \frac{dy}{e^y} = e^x dx \). Once in this form, each side of the equation only contains one variable, making the equation easier to solve through integration. This method is powerful because it converts complex differential equations into simpler integral forms.

Integration of Functions

Integration of functions is the process of finding a function whose derivative is given. In calculus, integration is the inverse operation to differentiation. When dealing with separable differential equations, once the equation is rewritten with separated variables, we integrate both sides to find the solution.

For example, in our exercise, we integrated \( -e^{-y} dy \) to give \( -e^{-y} \) and \( e^x dx \) to give \( e^x \). Integration helps us find a general formula based on the continuous sum of the infinitesimals represented by the differential equation.

Natural Logarithm

The natural logarithm is the logarithm to the base \( e \) (Euler's number, approximately 2.718281828459045). It is denoted by \( \ln x \) and is widely used in mathematics, physics, and engineering because of its simple derivative which is \( \frac{1}{x} \). The natural logarithm of a number is the power to which \( e \) must be raised to get that number.

In the initial value problem, applying the natural logarithm allows us to solve for \( y \) by utilizing the property that \( \ln(e^y) = y \) to isolate the variable \( y \) after integration.

Solving Differential Equations

Solving differential equations involves finding a function or a class of functions that satisfy the equation. Initial value problems, such as the one given in the exercise, require us to find a specific solution that also satisfies an initial condition. We usually start by isolating and integrating based on the type of differential equation we have. Then, we apply any given initial conditions to find the constants of integration, resulting in the specific solution that fits the problem's criteria.

In our example, after integrating and applying the initial condition \( y(0)=\ln 3 \), we found the constant of integration \( C_1 \), leading us to the particular solution that satisfies both the differential equation and the initial condition.