Question

Given that \(y_{1}\) is a solution of the given equation, use the method suggested by Exercise 55 to find other solutions. $$ x y^{\prime}=x^{3}+\left(1-2 x^{2}\right) y+x y^{2} ; \quad y_{1}=x $$

Short answer

Based on the given equation \(x y'=x^3+(1-2x^2)y+xy^2\) and the known solution \(y_1=x\), we found the general solution: $$ y(x) = x + \frac{2}{3}x^3 + x + \frac{1}{2}u^2x + C $$ where \(C\) is an arbitrary constant.

Step-by-step solution

Substitute \(y=y_1 + u\) in the given equation

Replace \(y\) by \(x + u(x)\) in the given equation to obtain the equation for \(u(x)\): $$ x(x+u)^\prime=x^3+(1-2x^2)(x+u)+x(x+u)^2 $$

Expand and simplify the equation

Expand the equation from Step 1 and simplify terms: \begin{align*} x\left(\frac{d(x+u)}{dx}\right) & =x^3+(1-2x^2)(x+u)+x(x+u)^2 \\ x(x'+u') &= x^3 + (1 - 2x^2)(x + u) + x(x^2 + 2xu + u^2) \\ x(x'+u') &= x^3 + x - 2x^3 + xu - 2x^2u + x^3 + 2x^2u + xu^2 \\ x(u') &= 2x^3 + x(1 + u + u^2) \\ \end{align*}

Divide by x and separate variables

Now, divide by \(x\) and separate the variables \(x\) and \(u\) to prepare for integration: $$ u' = 2x^2 + (1+u+u^2) $$

Integrate and find a general solution

Integrate both sides of equation from Step 3 with respect to \(x\): $$ \int \frac{du}{dx} \, dx = \int \left(2x^2+(1+u+u^2)\right) dx $$ Integrate both sides: $$ u(x) = \frac{2}{3}x^3 + x + \frac{1}{2}u^2x + C $$

Write the final solution

Writing the solution in terms of \(y\) by substituting \(y_1 + u\) for \(y\): $$ y = x + u(x) = x + \left(\frac{2}{3}x^3 + x + \frac{1}{2}u^2x + C\right) $$ Thus, the general solution is given by: $$ y(x) = x + \frac{2}{3}x^3 + x + \frac{1}{2}u^2x + C $$ with \(C\) being an arbitrary constant.

Key concepts

Method of Undetermined Coefficients

The Method of Undetermined Coefficients is a popular technique used in solving non-homogeneous linear differential equations. It involves guessing the form of a particular solution and determining the coefficients by plugging the guess into the equation. Here's an outline of how it generally works:

• Identify the type of function: Determine the type of non-homogeneous term (e.g., polynomial, exponential, trigonometric) present in the differential equation.
• Formulate a guess: Based on the identification, propose a form for the particular solution, often including unknown coefficients.
• Substitute and solve: Substitute the guess into the differential equation and solve for the undetermined coefficients. This often requires algebraic manipulation and comparison of like terms.

This method is effective for linear equations with constant coefficients and familiar functional forms, as they allow the equations to be simplified enough to solve for these coefficients easily. However, it can become cumbersome with complex or unconventional functions.

Integration Techniques

Integration is a fundamental technique used across mathematics, important for finding solutions to differential equations. Several methods are available, each suited to different types of integrands. Here are some common techniques:

• Substitution: This technique simplifies complex integrals by substituting a part of the integrand with a single variable, reducing the original integral into a basic form that's easier to solve.
• Integration by Parts: Useful for products of functions, it is derived from the product rule of differentiation and can break down the integrand into simpler parts.
• Partial Fractions: Applied when integrating rational functions, where the integrand is split into simpler fractions that can each be integrated individually.

Each of these methods is employed based on the structure of the integrand, allowing us to tackle more complex integrals by breaking them into manageable pieces. Mastery of these techniques is vital for solving differential equations as many solutions involve finding the integral of a derivative.

Separation of Variables

Separation of Variables is a powerful technique for solving certain types of differential equations where variables can be separated on opposite sides of the equation. Here's how it typically works:

• Identify separable form: The equation is of the form that can be rearranged to isolate one variable and its differential on one side, and another variable on the other side.
• Rearrange equation: Separate the differentials such that one side contains only functions of one variable and its differential (e.g., \( dy \) terms), and the other side contains the second variable (e.g., \( dx \) terms).
• Integrate both sides independently: Next, integrate both sides with respect to their own variables, typically leading to a couple of simpler integrals.

This method is particularly useful for first-order differential equations where isolating variables can transition a complex problem into more accessible integrations. It allows mathematicians and students to find solutions through direct integration, facilitating the derivation of explicit solutions and improving understanding of the underlying phenomena described by the equations.