Problem 54 - Differential Equations and Linear Algebra

Question

Use the result from the previous problem to solve the given differential equation. For Problem 54 impose the given initial condition as well. $$y^{\prime}=(9 x-y)^{2}, \quad y(0)=0$$

Step-by-Step Solution

Answer

The exact solution of the given differential equation \(y^{\prime} = (9x - y)^2\) with the initial condition \(y(0) = 0\) is: \(y(x) = \frac{9x^2 + 9x - 1}{x+1}\).

Step 1: Rearranging the differential equation

First, we'll rewrite the given differential equation as: \(\frac{dy}{dx} = (9x - y)^2\) Then, we'll rearrange the terms to get: \(\frac{dy}{(9x - y)^2} = dx\)

Step 2: Integration

Now, we'll integrate both sides of the equation: \(\int \frac{dy}{(9x - y)^2} = \int dx\)

Step 3: Substitution

Let's make a substitution to simplify the integral on the left side: Let \(u = 9x - y\), then \(\frac{du}{dy} = -1\), so \(dy = -du\). Now, our integral becomes: \(-\int \frac{du}{u^2} = \int dx\)

Step 4: Solving the integrals

Integration of both sides yields: \(\frac{1}{u} = x + C\) Note that we have only one constant of integration, as both sides were integrated.

Step 5: Back-substitution

Now, let's substitute back \(u = 9x - y\): \(\frac{1}{(9x - y)} = x + C\) To find the value of the constant C, we will use the initial condition \(y(0) = 0\): \(\frac{1}{(9(0) - 0)} = 0 + C\) This gives us \(C = 1\).

Step 6: Finalizing the solution

Substituting the value of C back into the equation: \(\frac{1}{(9x - y)} = x + 1\) Now, let's solve for y: \(1 = (9x - y)(x + 1)\) \(1 = 9x^2 + 9x - xy - y\) \(y(x+1) = 9x^2 + 9x - 1\) \(y = \frac{9x^2 + 9x - 1}{x+1}\) The exact solution of the given differential equation with the initial condition is: \(y(x) = \frac{9x^2 + 9x - 1}{x+1}\)

Key Concepts

Integration

Integration is a fundamental technique in calculus, used to find functions from their rates of change. In the context of differential equations, integration can help us to revert a differential equation back into a function.
To integrate a differential equation, recognize it as separating differentials. For example, starting with \( \frac{dy}{(9x - y)^2} = dx \), we can take the integral of both sides to begin solving for \(y\).
Integration aims to counteract differentiation, essentially 'undoing' it to provide the original function form.
In our case, integrating both sides step-by-step will eventually lead us to the general solution of the equation. However, when integrating, it's pivotal to remember to include the constant of integration \(C\).
Once integrated, solving such an integral involves handling initial conditions to find specific solutions.

Substitution

Substitution is a useful method to simplify integrals and differential equations by replacing variables. It’s especially handy when equations are complex or unwieldy.
In our problem, substitution helps in transforming the integral into a simpler form to solve.

• First, choose a substitution that makes the integral more manageable.
• For example, setting \(u = 9x - y\), changes the differential \(dy\) to \(-du\).
• This transforms our integral problem to \(-\int \frac{du}{u^2} = \int dx\), making it look easier to integrate.

After substituting, perform the integration. Later, remember to substitute back to the original variables to relate your calculations to the initial problem.
This method often simplifies complex integrations by reducing them to standard forms.

Initial Conditions

Initial conditions are additional information given for solving differential equations. They specify the value of the function or its derivatives at a particular point.
In this exercise, the initial condition is given as \(y(0) = 0\). This condition helps in determining the specific constant of integration \(C\), so you can find a unique and exact solution rather than a family of possible solutions.
To employ the initial condition, substitute \(x = 0\) and \(y = 0\) in the integrated result to solve for \(C\).
This process ensures that the solution not only satisfies the differential equation, but also matches the specific scenario or problem requirements.
Without applying initial conditions, solutions would exclude specific real-world or problem-specific insights.

Exact Solution

An exact solution to a differential equation is a precise expression for the function that satisfies both the equation itself and any imposed initial conditions.
In this case, the solution is found as \(y(x) = \frac{9x^2 + 9x - 1}{x+1}\), which is derived by integrating the rearranged equation and applying the initial conditions.

• The solution process involved transforming the differential equation, integrating with respect to appropriate variables, and simplifying through substitution.
• Finally, using the initial condition \(y(0) = 0\), the exact form of \(y(x)\) was determined by solving for the constant \(C\) and back-substituting variables involved.

An exact solution is crucial because it provides precise modeling of phenomena described by the differential equation while adhering to initial conditions—it is the unique solution needed by the problem posed.