Problem 26 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$\frac{d y}{d x}=\frac{2 x-y}{x+4 y}, \quad y(1)=1$$

Step-by-Step Solution

Answer

The implicit solution to the given initial-value problem is: $$-\ln|2x - y| = x + 4y - 5$$

Step 1: Identify the type of differential equation

This is a first-order, separable differential equation. The aim is to separate the variables, y and x, in the equation, and then integrate both sides independently.

Step 2: Separate variables

We want to rewrite the given equation in the form P(x)dx = Q(y)dy. From the given equation, $$\frac{dy}{dx}=\frac{2x-y}{x+4y}$$ Multiply both sides by \(dx\), and then arrange all terms containing y on the left and all terms containing x on the right: $$\frac{dy}{2x-y} = \frac{dx}{x+4y}$$ Now we have separated the variables.

Step 3: Integrate both sides

Next, we integrate both sides of the equation with respect to their respective variables: $$\int \frac{dy}{2x - y} = \int \frac{dx}{x + 4y}$$ In order to integrate, we need to make a substitution for both sides. Let \(v = 2x - y\) for the left integral and let \(u = x + 4y\) for the right integral. For the left side, we get \(dv = -dy\) after taking the differential w.r.t y. The left integral becomes: $$-\int \frac{dv}{v} = -\ln|v|+C_1$$ For the right side, we get \(du = (1+4\frac{dy}{dx})dx\), and so \(\frac{dy}{dx} = \frac{du/dx - 1}{4}\). Replacing the given differential equation, we get \(\frac{du/dx - 1}{4} = \frac{1}{u}\). Now we can solve for du/dx and integrate the right side with respect to x: $$ \frac{du}{dx} = 1 + \frac{4}{u}$$ $$\int \frac{du}{1 + \frac{4}{u}} dx = \int dx$$ $$\int u \frac{du}{u + 4} = x + C_2$$

Step 4: Substitute back to x and y

Now we can write the solutions in terms of x and y, by substituting back from the definitions of u and v: $$-\ln|2x - y| = x + 4y + C$$ where \(C = C_2 -C_1\) is a new integration constant.

Step 5: Use the initial condition to find the constant

Using the initial condition, y(1) = 1, substitute x=1 and y=1 into the equation above to find the constant C: $$-\ln|2(1) - 1| = 1 + 4(1) + C$$ $$-\ln{1} = 5 + C$$ $$C = -5$$

Step 6: Write the final solution

Finally, we substitute the found constant C back into the solution and rewrite the equation in the original variables: $$-\ln|2x - y| = x + 4y - 5$$ This is the implicit solution to the given initial-value problem.

Key Concepts

Initial-value problem

An initial-value problem combines a differential equation with a specific condition that the solution must satisfy, such as a value at a particular point. In our example, we have the differential equation \( \frac{d y}{d x} = \frac{2 x - y}{x + 4 y} \), along with the initial condition \( y(1) = 1 \). This condition provides additional information that is crucial for finding a unique solution.

The initial condition essentially tells us that the curve, which represents the solution to the differential equation, must pass through the point (1, 1). Solving an initial-value problem usually involves:

• Identifying the type of differential equation involved.
• When possible, finding a general solution to the differential equation.
• Substituting the initial values given to find specific values for any constants involved, which ensures that the solution satisfies the initial condition.

By using the initial condition \( y(1) = 1 \), we find a unique solution that fits both the differential equation and this specific condition.

First-order differential equations

A first-order differential equation involves the first derivative of a function but not higher derivatives. In other words, it contains terms involving \( \frac{d y}{d x} \) but not \( \frac{d^2 y}{d x^2} \) or other higher-order terms. Our given problem \( \frac{d y}{d x} = \frac{2 x - y}{x + 4 y} \) is a perfect example.

First-order differential equations are fundamental because they often appear in real-world applications, like modeling population growth or cooling processes. They come in various forms, such as separable, linear, and exact equations. Understanding these classifications helps in determining the best approach for solving them.

To solve them:

• Identify the structure of the differential equation (e.g., linear, separable).
• Apply the appropriate method to find a general solution.
• Use additional conditions, if any, to determine specific solutions.

First-order differential equations represent a core part of calculus and are essential for both theoretical studies and practical applications.

Separable differential equations

Separable differential equations are a type of first-order differential equation that can be rearranged so that all terms involving one variable are on one side of the equation and all terms involving the other variable are on the opposite side. This makes them particularly straightforward to solve by integration.

Consider the equation: \( \frac{d y}{d x} = \frac{2 x - y}{x + 4 y} \). By re-arranging, we derive:

• \( \frac{dy}{2x-y} = \frac{dx}{x+4y} \) - where each side of the equation contains only one variable.

We can then integrate both sides separately. This process is as simple as reversing the differentiation, allowing us to find an explicit or implicit solution.

Tips for solving separable differential equations include:

• Carefully manipulating the equation to separate variables completely.
• Conducting integration for both sides of the equation.
• Ensuring any constants of integration are included during integration.

Separable differential equations are widely used because they simplify into a series of straightforward steps, making them accessible even for those new to differential equations.