Question

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

Short answer

Question: Find other solutions to the given differential equation: \(x\frac{dy}{dx}=2 -x + (2x - 2)y - xy^2\) if \(y_1 = 1\). Answer: The new solution \(y_2(x)\) for the differential equation is given by: $$ y_{2}(x) = 1 - \frac{e^{(x^2-2x)}}{\int -e^{(x^2-2x)}dx + C} $$ where the integral of the exponential function and an arbitrary constant \(C\) determine the specific solution.

Step-by-step solution

Write down the relationship between \(y_{1}\) and \(y_{2}\)

Since \(y_1=1\) and we are looking for a new solution \(y_2\), we use a relationship \(y_2=1+u(x)\), where \(u(x)\) is an arbitrary function of \(x\).

Calculate the derivative of \(y_2\)

Find the derivative of \(y_2\): $$ y^{\prime}_{2} = \frac{d}{dx} (1 + u(x)) = u'(x) $$

Substitute \(y_2\) and \(y_{2}'\) into the given equation

Plug the expressions for \(y_2\) and its derivative into the original differential equation: $$ x\cdot u^{\prime}(x) = 2-x+(2x-2)(1+u(x))-x(1+u(x))^2 $$

Simplify the equation for \(u(x)\)

Simplify the equation obtained in Step 3, canceling the terms that do not contain \(u(x)\) or \(u^{\prime}(x)\) $$ x\cdot u^{\prime}(x) = (2x-2)u(x)-xu^{2}(x) $$

Solve the equation for \(u(x)\)

To solve this new equation for \(u(x)\), we can try the method of integrating factors, or we can attempt to separate the variables and then integrate both sides. The equation we have is a Bernoulli equation: $$ x\cdot u'(x)+ (2x - 2)u(x) = xu^{2}(x) $$ Divide through by \(xu^{2}(x)\): $$ -\frac{1}{u^2(x)}u'(x) +\frac{(2x-2)}{x} \frac{-1}{u(x)} = -1 $$ Now we substitute \(w(x)=-\frac{1}{u(x)}\) and \(w'(x)=\frac{u'(x)}{u^{2}(x)}\). The equation becomes: $$ -w'(x) + (2x-2)w(x) = -1 $$ Now we can use the integrating factor method. The integrating factor is: $$ \rho(x)=e^{\int (2x-2) dx} = e^{(x^2-2x)} $$ Now we multiply the whole equation by the integrating factor \(\rho(x)\): $$ e^{(x^2-2x)}(-w'(x) + (2x-2)w(x)) = - e^{(x^2-2x)} $$ Now the left side becomes the derivative of the product \(\rho(x)w(x)\): $$ \frac{d}{dx}(e^{(x^2-2x)}w(x))= -e^{(x^2-2x)} $$ Integrate both sides with respect to x: $$ e^{(x^2-2x)}w(x)) = \int -e^{(x^2-2x)}dx + C $$ Now solve for \(w(x)\) and then substitute back for \(u(x)\): $$ w(x)= \frac{\int -e^{(x^2-2x)}dx + C}{e^{(x^2-2x)}} $$ $$ -\frac{1}{u(x)} =\frac{\int -e^{(x^2-2x)}dx + C}{e^{(x^2-2x)}} $$ $$ u(x) = -\frac{e^{(x^2-2x)}}{\int -e^{(x^2-2x)}dx + C} $$

Get the new solution \(y_{2}\)

Since \(y_{2} = 1 + u(x)\), we get the new solution as: $$ y_{2}(x) = 1 - \frac{e^{(x^2-2x)}}{\int -e^{(x^2-2x)}dx + C} $$ The new solution \(y_2\) depends on the integral of exponential function and an arbitrary constant \(C\).

Key concepts

Method of Integrating Factors

The "Method of Integrating Factors" is a powerful technique used to solve certain kinds of first-order linear differential equations. Specifically, it deals with equations of the form \(y' + P(x)y = Q(x)\). The method relies on calculating an integrating factor, \(\mu(x)\), which transforms the equation into one that can be integrated directly.

Here's the basic process:

• Identify \(P(x)\) in the equation, which is the coefficient of \(y\).
• Compute the integrating factor \(\mu(x) = e^{\int P(x) \, dx}\).
• Multiply the entire differential equation by this integrating factor.
• The left-hand side should now be the derivative of \(\mu(x)y\).
• Integrate both sides to find \(y\).

In this exercise, the method of integrating factors helps simplify the equation, allowing us to solve for a new function \(w(x)\). The solution is then found by undoing the substitution made initially.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are crucial in modeling and understanding various physical, economic, and biological processes. In simpler terms, they demonstrate how a particular quantity changes with respect to another quantity, often time.

There are several types of differential equations:

• Ordinary Differential Equations (ODEs): Involve functions of a single variable and their derivatives.
• Partial Differential Equations (PDEs): Involve multiple independent variables.

The exercise involves an ordinary differential equation given by \(x y^{\prime} = 2-x+(2x-2)y-xy^{2}\). Solving this equation involves finding a function \(y(x)\) that satisfies this relationship. The process often requires manipulating and simplifying the equation, possibly using techniques like substitution or integrating factors.

Integration

Integration is the process of finding the integral of a function, which is essentially the reverse of differentiation. It is a fundamental operation in calculus, used for computing areas, volumes, and other quantities that accumulate.

Key concepts in integration include:

• The Indefinite Integral: Represents a family of functions \(F(x)\) that differentiate to \(f(x)\), written as \( \int f(x) \, dx = F(x) + C\).
• The Definite Integral: Provides the accumulated quantity of \(f(x)\) over an interval \([a, b]\) and is denoted by \( \int_{a}^{b} f(x) \, dx \).
• Techniques of Integration: Methods like substitution, integration by parts, and partial fraction decomposition help solve complex integrals.

In the given step-by-step solution, integration is used to solve the differential equation once it has been suitably transformed. Typically, calculating the integral of a function with an integrating factor reveals the necessary relationships to determine the unknown function \(y(x)\).