Question

Find the solution to the initial-value problem.\(y^{\prime}=e^{y} 5^{x}, y(0)=\ln (\ln (5))\)

Short answer

The solution is \( y = -\ln\left(-\frac{5^x}{\ln(5)} + \ln(5) + \frac{1}{\ln(5)} \right). \)

Step-by-step solution

Set Up the Differential Equation

To solve the differential equation, begin by substituting the expression for the given derivative: \[ y' = e^y 5^x. \] This equation states that the derivative of \( y \) with respect to \( x \) is equal to \( e^y \cdot 5^x \).

Separate Variables

Next, rearrange the equation to separate the variables \( y \) and \( x \). Divide both sides by \( e^y \):\[ \frac{dy}{e^y} = 5^x dx. \] This separates terms involving \( y \) on one side and terms involving \( x \) on the other.

Integrate Both Sides

To find \( y \), integrate both sides. The integral of the left side is:\[ \int \frac{dy}{e^y} = \int e^{-y} dy = -e^{-y}. \]The integral of the right side is:\[ \int 5^x dx = \frac{5^x}{\ln(5)}. \] Thus, we integrate to get:\[ -e^{-y} = \frac{5^x}{\ln(5)} + C \] where \( C \) is the constant of integration.

Solve for y

To express \( y \) explicitly, solve the integrated equation for \( y \):Subtract \( \frac{5^x}{\ln(5)} \) from both sides:\[ -e^{-y} = \frac{5^x}{\ln(5)} + C. \]Multiply through by \(-1\):\[ e^{-y} = -\frac{5^x}{\ln(5)} - C. \]Take the natural logarithm to solve for \( y \):\[ y = -\ln\left(-\frac{5^x}{\ln(5)} - C\right). \]

Apply Initial Condition

Use the initial condition \( y(0) = \ln(\ln(5)) \) to solve for \( C \). Substituting \( x = 0 \),\( y = \ln(\ln(5)) \) into the equation gives:\[ e^{-\ln(\ln(5))} = -\frac{1}{\ln(5)} - C. \]Plug in the value for \( e^{y} = \frac{1}{\ln(5)} \):\[ \ln(5) = -\frac{1}{\ln(5)} - C. \]Solve for \( C \):\[ C = -\ln(5) - \frac{1}{\ln(5)}. \]

Substitute Back to Get Final Solution

Substitute \( C \) back into the equation obtained in Step 4 to get the solution:\[ y = -\ln\left(-\frac{5^x}{\ln(5)} + \ln(5) + \frac{1}{\ln(5)} \right). \]This is the particular solution to the initial-value problem.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. These equations often model real-world phenomena, such as population growth, motion, or electrical circuits.
Differential equations can be ordinary (ODE) or partial (PDE):

• Ordinary when they involve a single variable.
• Partial when they involve multiple variables.

In the context of an initial-value problem, you're given a differential equation along with a specified value of the function at a certain point.
This additional piece of information allows you to determine a unique solution. In our example, the differential equation is given in the form \( y' = e^y 5^x \), where the derivative of \( y \) with respect to \( x \) is expressed in terms of \( y \) and \( x \). Knowing the initial condition \( y(0) = \ln(\ln(5)) \), helps us find the precise solution.

Integration

Integration is the process of finding the antiderivative or integral of a function. It's the inverse operation of differentiation and is crucial in solving differential equations.
When we integrate both sides of the rearranged differential equation \( \frac{dy}{e^y} = 5^x dx \), we're finding the original functions that correspond to these derivatives.
The integration process involves solving:

• Left side: \( \int e^{-y} \, dy = -e^{-y} \)
• Right side: \( \int 5^x \, dx = \frac{5^x}{\ln(5)} \)

These integrations help us move closer to a solution by expressing the relationship between \( y \) and \( x \). Each function yields a constant of integration, \( C \), because the integral of a function is not unique without further information, such as an initial condition.

Separation of Variables

Separation of variables is a key technique for solving differential equations. It involves rearranging the equation so that each variable and its derivative are on opposite sides. This allows each side to be integrated individually.
With our differential equation, we move all terms involving \( y \) to one side and terms involving \( x \) to the other.
This process yields the equation: \( \frac{dy}{e^y} = 5^x \, dx \).

• This method simplifies the equation and prepares it for integration.
• Each side can be integrated independently.

Separation of variables is particularly useful for first-order differential equations, where the solution involves exponential functions, as seen in our example.

Exponential Functions

Exponential functions, such as \( e^y \) and \( 5^x \), appear frequently in the study of differential equations. They describe processes involving growth or decay at a constant rate.
The general form is \( a^{x} \), where \( a \) is a constant. In many mathematical models, including our example, exponential functions represent rapid changes.
The solution to our initial-value problem involves reversing these exponential forms through integration.

• Through separation of variables, we isolate and integrate these exponential parts.
• Using the initial condition helps us find the specific form of the solution.

In our exercise, solving the initial-value problem results in a detailed expression involving logarithms, which stem from handling exponential functions during the solving process.