Question

Solve the given differential equations. $$ x^{4} \frac{d y}{d x}+2 x^{3} y=1 $$

Short answer

The solution to the given differential equation is: \( y(x) = \frac{1}{3x^2} + \frac{C}{x^5} \).

Step-by-step solution

Identify a(x), b(x), and c(x)

Given the equation, \( x^{4} \frac{d y}{d x}+2 x^{3} y=1 \) Let's identify the functions a(x), b(x), and c(x) in the standard form: \( a(x) = x^4 \) \( b(x) = 2x^3 \) \( c(x) = 1 \)

Find the Integrating Factor (IF)

To find the integrating factor (IF), first calculate: \( r(x) = \int \frac{b(x)}{a(x)} dx = \int \frac{2x^3}{x^4} dx \) Solve the integral r(x): \( r(x) = \int \frac{2x^3}{x^4} dx = \int 2x^{-1} dx = 2\int x^{-1} dx = 2(\ln|x| + C) \) Now, the integrating factor (IF) is given by: \( IF = e^{r(x)} = e^{2(\ln|x| + C)}=x^2 e^{2C} \) Since C is an arbitrary constant, we can absorb the exponential e^{2C} into the constant, i.e., e^{2C} = K, where K is an arbitrary constant. Therefore, \( IF = Kx^2 \) For simplicity, let's take K = 1: \( IF = x^2 \)

Multiply the equation by the Integrating Factor

Now, we will multiply our given differential equation by the integrating factor: \( x^{4} \frac{d y}{d x} + 2x^3 y = 1 \) Multiply by IF: \( x^2(x^{4} \frac{d y}{d x} + 2x^3 y) = x^2 \) Simplify the equation: \( x^6 \frac{d y}{d x} + 2x^5 y = x^2 \)

Integrate both sides of the equation

Integration of the left side of the equation is achieved using the product rule of differentiation in reverse: \( \frac{d}{dx}(x^5y) = x^6 \frac{d y}{d x} + 2x^5 y \) Next, integrate both sides with respect to x: \( \int (x^6 \frac{d y}{d x} + 2x^5 y) dx = \int x^2 dx \) On the left side, apply the reverse product rule: \( x^5 y = \int x^2 dx \) Now, integrate the right side: \( x^5 y = \frac{1}{3}x^3 + C \)

Solve for y(x)

Finally, we need to solve for y(x): \( y(x) = \frac{1}{x^5}(\frac{1}{3}x^3 + C) \) \( y(x) = \frac{1}{3x^2} + \frac{C}{x^5} \)

Final Answer

The solution to the given differential equation is: \( y(x) = \frac{1}{3x^2} + \frac{C}{x^5} \)

Key concepts

Integrating Factor

The concept of an integrating factor is crucial when dealing with linear ordinary differential equations. An integrating factor is a function that, when multiplied with the entire differential equation, simplifies the equation into a form that is more easily solvable by integration.To find the integrating factor for a differential equation, say in the form of \( a(x) \frac{d y}{d x} + b(x) y = c(x) \), we first determine the function \( r(x) = \int \frac{b(x)}{a(x)} dx \). This integral helps us determine the integrating factor:

• Calculate the integral \( r(x) = \int \frac{b(x)}{a(x)} dx \).
• The integrating factor \( IF \) is then given by \( e^{r(x)} \).
• Multiplying the entire differential equation by this \( IF \) transforms it into an easily integrable form.

For the exercise, the integrating factor simplified the differential equation and enabled step-by-step resolution by turning the left side into a perfect product of derivatives. This is a powerful tool in solving such equations.

Product Rule

The product rule is a fundamental concept in calculus used to differentiate products of two functions. It states that the derivative of the product of two functions, say \( u(x) \) and \( v(x) \), is given by:\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]In differential equations, the reverse process is often used to integrate expressions that match this form.In the exercise, after multiplying by the integrating factor, the equation was manipulated to take the form of a derivative of a product. Specifically, we recognize the left hand side as the derivative of \( x^5y \), which allowed us to proceed with integration more straightforwardly. This reversal of the product rule, seeing the sum of derivatives as a single product's derivative, simplifies and unlocks the integration process for solving many differential equations.

Ordinary Differential Equation

An ordinary differential equation or ODE, is a mathematical equation that relates some function with its derivatives. ODEs are broadly classified into linear and nonlinear equations depending on the degree of the variables and their derivatives. The problem in the exercise is a first-order linear differential equation. In such equations, the derivatives of the function appear to the first degree, making them suitable for being solved using integrating factors among other methods. Techniques to tackle these equations, including finding an integrating factor, converting them into a form with constant coefficients, or even solving them directly through known functions like exponentials and polynomials, are foundational skills in mathematics and physics. Understanding ODEs is essential for modelling real-world phenomena where rates of change are involved, such as in engineering or natural sciences.

Integration Techniques

The exercise employs several integration techniques, combining them to work through the problem efficiently. Integration is the process of finding the integral, or the reverse of differentiation, which often involves recognizing a specific pattern or using a strategic method to solve integrals of complex expressions.Some key integration techniques involved include:

• Basic Power Rule: Used when integrating terms like \( x^n \), involving \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
• Integration by Substitution: While not explicitly detailed, it's often helpful in simplifying complex terms for integration.
• Using the product rule in reverse: Recognizes the form \( u'(x)v(x) + u(x)v'(x) \) as a derivative product, which can simplify integration.

Understanding how to apply these techniques interchangeably is crucial for handling different types of integrals found in differential equation solutions.