Problem 7 - Advanced Engineering Mathematics

Question

In Problems 1-22, solve the given differential equation by separation of variables. $$ \frac{d y}{d x}=e^{3 x+2 y} $$

Step-by-Step Solution

Answer

The solution is \(y = -\frac{1}{2}\ln\left(-\frac{2}{3}e^{3x} - 2C\right)\).

Step 1: Set up the separation

To solve the differential equation \(\frac{d y}{d x}=e^{3 x+2 y}\), we need to separate the variables so that all terms involving \(y\) are on one side of the equation and all terms involving \(x\) are on the other. Let's express the equation in the following form: \(\frac{d y}{e^{2y}} = e^{3x} \, d x\).

Step 2: Integrate both sides

Now, integrate both sides of the equation. For the left side, integrate with respect to \(y\): \(\int \frac{1}{e^{2y}} \, d y\), and for the right side, integrate with respect to \(x\): \(\int e^{3x} \, d x\).

Step 3: Solve the integrals

The integral of the left side is \(\int e^{-2y} \, d y = -\frac{1}{2}e^{-2y} + C_1\), and the integral of the right side is \(\int e^{3x} \, d x = \frac{1}{3}e^{3x} + C_2\).

Step 4: Combine the solutions

Equating the results of the two integrals, we have: \(-\frac{1}{2}e^{-2y} = \frac{1}{3}e^{3x} + C\), where \(C = C_2 - C_1\).

Step 5: Solve for \(y\)

Isolate \(y\) in the equation. Start by multiplying through by -2 to get \(e^{-2y} = -\frac{2}{3}e^{3x} - 2C\). Then, take the natural logarithm to solve for \(y\): \(-2y = \ln\left(-\frac{2}{3}e^{3x} - 2C\right)\), leading to \(y = -\frac{1}{2}\ln\left(-\frac{2}{3}e^{3x} - 2C\right)\).

Key Concepts

Separation of Variables

Separation of variables is a method used to solve first-order differential equations. The goal is to rearrange the equation so that each variable and its differential are on opposite sides of the equation. This allows us to integrate both sides independently.

In the provided problem, we have:
\(\frac{dy}{dx} = e^{3x+2y}\).
Our task is to separate the terms involving \(y\) and its derivative \(dy\) from those involving \(x\) and \(dx\). We achieve this by rewriting the equation as:

• \(\frac{dy}{e^{2y}} = e^{3x} \, dx\)

Now, the left side of the equation contains all the \(y\) terms with \(dy\), while the right contains all \(x\) terms with \(dx\).

The goal of separation of variables is to make the integration process straightforward since we can integrate each side with respect to its respective variable. This technique is particularly powerful because it simplifies the differential equation into two manageable integrals.

Integration Methods

Integration is a fundamental tool when dealing with differential equations. Once the variables are separated, the next step is to integrate each side of the equation with respect to its variable.

In our exercise, after separating variables, the equation becomes:

• \(\int \frac{1}{e^{2y}} \, dy = \int e^{3x} \, dx\)

For the left-hand side, you integrate with respect to \(y\). The integral of \(e^{-2y}\) involves using an exponential rule for integrating functions of the form \(e^{ay}\), resulting in:

• \(-\frac{1}{2}e^{-2y} + C_1\)

For the right-hand side, integrating \(e^{3x}\) with respect to \(x\) also uses an exponential rule, giving:

• \(\frac{1}{3}e^{3x} + C_2\)

The result of these integrations allows you to form an equation that can be solved further. It’s crucial to understand integration techniques such as substitution and recognizing standard integrals to simplify the solving process.

Solving First-Order Differential Equations

To solve a first-order differential equation, especially one separable by variables, involves combining the methods of separation and integration. After separating the variables and integrating, you typically have an expression that includes the integration constants, which need to be managed properly.

In the problem's solution, after the integrations, we arrived at:

• \(-\frac{1}{2}e^{-2y} = \frac{1}{3}e^{3x} + C\)

where \(C = C_2 - C_1\).
Here, the task is to solve for \(y\) explicitly. This may involve algebraic manipulation, like multiplying through by a factor or using logarithmic properties.

For this equation, we multiply by \(-2\) to simplify:

• \(e^{-2y} = -\frac{2}{3}e^{3x} - 2C\)

Then, applying logarithms to both sides helps isolate \(y\):

• \(-2y = \ln\left(-\frac{2}{3}e^{3x} - 2C\right)\)

Solving for \(y\), we finally get:

• \(y = -\frac{1}{2}\ln\left(-\frac{2}{3}e^{3x} - 2C\right)\)

The solutions to such equations often include an arbitrary constant, reflecting the presence of a family of solutions, highlighting the role of initial conditions in differential equations.