Problem 9

Question

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ x \frac{d y}{d x}-y=x^{2} \sin x $$

Step-by-Step Solution

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Answer

The general solution is $ y = -x \cos x + Cx $, valid for $(- fty, 0) \cup (0, \infty)$, with no transient terms.

1Step 1: Identify Type of Differential Equation

The given differential equation is $ x \frac{d y}{d x} - y = x^2 \sin x $. This is a linear first-order differential equation.

2Step 2: Standard Form of Linear Equation

Rewrite the equation in the standard form $ \frac{d y}{d x} + P(x) y = Q(x) $ by dividing the entire equation by $ x $ (assuming $ x eq 0 $): \[ \frac{dy}{dx} - \frac{1}{x} y = x \sin x. \] Thus, $ P(x) = -\frac{1}{x} $ and $ Q(x) = x \sin x $.

3Step 3: Determine the Integrating Factor

The integrating factor $ \mu(x) $ is given by $ e^{\int P(x) \, dx} $. Compute $ \int -\frac{1}{x} \, dx = -\ln |x| $, so $ \mu(x) = e^{-\ln |x|} = \frac{1}{x} $.

4Step 4: Multiply Through by the Integrating Factor

Multiplying the entire differential equation by the integrating factor $ \frac{1}{x} $ yields \[ \frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = \sin x. \]

5Step 5: Solve the Modified Equation

The left-hand side can now be rewritten as $ \frac{d}{dx} \left( \frac{y}{x} \right) $. Thus, \[ \frac{d}{dx} \left( \frac{y}{x} \right) = \sin x. \] Integrate both sides with respect to $ x $: \[ \frac{y}{x} = -\cos x + C, \] where $ C $ is the integration constant.

6Step 6: Solve for y

Multiply through by $ x $ to obtain $ y = -x \cos x + Cx $. This is the general solution.

7Step 7: Determine the Interval of Validity

The solution requires $ x eq 0 $ because of division by $ x $ earlier. So the largest interval over which the solution is defined is $ (-\infty, 0) \cup (0, \infty) $.

8Step 8: Identify Transient Terms

Transient terms often vanish as $ x \to \infty $ or $ x \to -\infty $. The term $ -x \cos x $ oscillates indefinitely and does not vanish. Therefore, there are no transient terms in the general solution.

Key Concepts

Linear differential equationsIntegrating factor methodGeneral solutionInterval of validity

Linear differential equations

A linear differential equation is an equation involving a function and its derivatives, where each term is either a constant or the product of a constant and the first power of the function or its derivatives.
This makes them much simpler compared to non-linear differential equations. Linear equations can be found in many dynamics or physical systems such as electrical circuits or heat conduction.
For first-order linear differential equations, the general form is expressed as:

$ \frac{dy}{dx} + P(x)y = Q(x) $

Where:

$ P(x) $ and $ Q(x) $ are functions of $ x $.

Integrating factor method

The integrating factor method is a crucial technique used to solve linear first-order differential equations.
It involves finding an integrating factor, which simplifies the equation to a form that can be easily integrated.
This factor helps in rewriting the left-hand side as a derivative of a product.Here’s how it works:

Consider an equation in the standard form: $ \frac{dy}{dx} + P(x)y = Q(x) $
The integrating factor, $ \mu(x) $, is given by $ e^{\int P(x) \, dx} $
Multiplying through the entire equation by $ \mu(x) $ transforms the equation.

This transformation allows you to integrate straightforwardly, simplifying solving the differential equation.

General solution

The general solution of a differential equation encompasses all possible solutions and includes an arbitrary constant, which adjusts based on initial conditions.
For a first-order differential equation, the solution provides a family of functions. In our case, the general solution is:

$ y = -x \cos x + Cx $

Where:

$ C $ is the constant of integration, representing an infinite number of solutions.

This general solution satisfies the original differential equation across its domain but depends on specific initial conditions to find the particular solution.

Interval of validity

The interval of validity refers to the range over which the solution to a differential equation remains valid.
It is the set of values for which the solution is well-defined and continuous.For the given solution:

The restriction $ x eq 0 $ arises due to division by $ x $ during the solving process, making $ x = 0 $ undefined.

Thus, the largest interval of validity here is:

$ (-\infty, 0) \cup (0, \infty) $

This means the solution is valid for all real numbers except zero.

Problem 9

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