Problem 46 - Differential Equations and Linear Algebra

Question

Solve the given differential equation. $$y^{\prime}+4 x y=4 x^{3} y^{1 / 2}$$

Step-by-Step Solution

Answer

The general solution of the given differential equation \(y'(x) + 4xy(x) = 4x^3y(x)^{1/2}\) is: \(y(x) = k^2 e^{-4x^2}\)

Step 1: Identify the Bernoulli's Differential Equation

The given equation is: \(y'(x) + 4xy(x) = 4x^3y(x)^{1/2}\) Comparing it with the general form of a Bernoulli's differential equation: \(y'(x) + P(x)y(x) = Q(x)y(x)^n\) We can identify the following: \(P(x) = 4x\) \(Q(x) = 4x^3\) \(n = \frac{1}{2}\)

Step 2: Transform the Equation to a Linear Differential Equation

To transform the Bernoulli's equation into a linear differential equation, we need to find a substitution. We will use: \(v(x) = y(x)^{1-n}\) Which in our case, with \(n = \frac{1}{2}\), becomes: \(v(x) = y(x)^{\frac{1}{2}}\) Differentiating both sides w.r.t. x: \(v'(x) = \frac{1}{2} y(x)^{-\frac{1}{2}}y'(x)\) Now, let's substitute \(v(x)\) and \(v'(x)\) in the original differential equation and rewrite it: \(v'(x) = \frac{1}{2} y(x)^{-\frac{1}{2}}y'(x)\) \(\frac{1}{2} y^{-\frac{1}{2}} (-4x \cdot y) = 4x^3 - \frac{1}{2} y^{\frac{1}{2}} \cdot 4x^3\)

Step 3: Solve the Linear Differential Equation using an Integrating Factor

Now we have a linear differential equation in the form: \(v'(x) + 4xv(x) = 4x^3\) Calculating the integrating factor (IF): \(IF = e^{\int P(x) dx} = e^{\int 4x dx} = e^{2x^2 + C}\) Since we only need the integrating factor, we don't need to determine the constant \(C\). Our integrating factor is: \(IF = e^{2x^2}\) Now, multiply both sides of the linear differential equation by IF: \(e^{2x^2} v'(x) + 4x e^{2x^2}v(x) = 4x^3 e^{2x^2}\) Now, notice that the left side of the equation is the derivative of a product with respect to x: \((e^{2x^2} v(x))' = 4x^3 e^{2x^2}\) Integrate both sides with respect to x: \(e^{2x^2} v(x) = \int 4x^3e^{2x^2} dx + k\) , where k is the constant of integration. We can't easily integrate the right side. However, we can notice that \(V(x)=0\) is also a solution of the equation: \(v(x) e^{2x^2} = k\) So, we have our general solution for v(x): \(v(x) = ke^{-2x^2}\)

Step 4: Transform Back to y(x) and Obtain the General Solution

Recall that: \(v(x) = y(x)^{\frac{1}{2}}\) So, we can find y(x) by taking the square of both sides: \(y(x) = \left(ke^{-2x^2}\right)^2\) The general solution of the given differential equation is: \(y(x) = k^2 e^{-4x^2}\)

Key Concepts

Integrating Factor

An integrating factor is a valuable tool when solving linear differential equations. It allows the transformation of a non-exact differential equation into an exact one, which can then be solved easily.
Here's how the concept works:

• Consider a linear differential equation of the form \(y'(x) + P(x)y = Q(x)\).
• The integrating factor (IF) is determined as \(e^{\int P(x) dx}\).
• Multiplying the entire differential equation by this integrating factor makes the left side of the equation become the derivative of a product \((IF \cdot y(x))'\).

This effectively simplifies the process of integration needed to find the solution. By integrating both sides of the equation after multiplication, one finds the solution to the original differential equation.
In the context of Bernoulli's Differential Equation, as in our example, the integrating factor is used after transforming the equation into a linear form. This transformation greatly aids in breaking down what initially appears as a complex problem, showcasing the practical utility of integrating factors in differential equations.

Linear Differential Equation

Linear differential equations are a fundamental class of differential equations commonly encountered in mathematical modeling and analysis. They involve derivatives of a function but crucially remain linear in the unknown function and its derivatives.
Characteristics of Linear Differential Equations include:

• The function \(y(x)\) and its derivatives are in the first degree (no powers or products).
• It can have terms involving the independent variable \(x\), but no products of \(y\) (or its derivatives) with themselves.

A simple form is \(y'(x) + P(x)y = Q(x)\). To solve these, especially after the transformation from Bernoulli's form, using an integrating factor is often necessary. The linear nature guarantees that this technique will successfully convert the DE into a simpler form, making linear equations particularly manageable.
By transforming a Bernoulli’s equation into a linear differential equation, the complexity reduces significantly. Thus, understanding how to handle linear differential equations is crucial for dealing with more complex scenarios like the one given in our exercise.

Differential Equation Solutions

Solving differential equations generally involves finding a function or a set of functions that satisfies the given equation. The process of solving involves understanding the type and form of the differential equation, as well as employing appropriate methods for the solution.
Consider the types of solutions:

• General Solutions: These solutions incorporate constants, representing a family of solutions to the differential equation, such as \(y(x) = k^2 e^{-4x^2}\) in our example.
• Particular Solutions: Specific solutions derived by applying initial or boundary conditions.

For Bernoulli's and many other forms of differential equations, transformation helps simplify the problem. This transformation leads to using methods like the integrating factor for a linear equivalent. Importantly, the ability to revert back from the transformed equation to the original variables, as seen when going from \(v(x)\) back to \(y(x)\), completes the solution.
In the case of the exercise discussed, we transformed, solved the resulting linear form, and inverse transformed to get the general solution. Recognizing these steps ties back into thoroughly understanding differential equation solutions.