Problem 12 Identify each of the differentia... [FREE SOLUTION]

Chapter 8: Problem 12

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x y^{\prime \prime}+y^{\prime}=4 x$$

Short Answer

Expert verified

Second-order linear; Solution: $y = x^2 + C \reit \reit D$.

Step by step solution

Identify the Type of Differential Equation

The given differential equation is $x y^{\treble}+y^{\treble}=4 x$. To identify its type, observe the order and linearity. This is a second-order linear differential equation because the highest derivative is second-order (\reit^2) and it can be written in the form $a_2(x) y^{\reit^}+ a_1(x) y^{\reit}+ a_0(x) y = g(x)$. Here, $a_2(x) = x$, $a_1(x) = 1$, $a_0(x) = 0$, and $g(x) = 4x$.

Standard Form

The standard form of a second-order linear differential equation is $a_2(x) y^{\reit^2} + a_1(x) y^{\reit} + a_0(x) y = g(x)$. Rewriting the given equation: $x y^{\riet^2}+ y^{\reit} = 4x$.

Reducing Order

To solve this, first reduce the order by substituting $y' = z$ which implies $y'' = z'$. Therefore, the equation becomes $x z' + z = 4x$.

Solve the First Order

The equation $x z' + z = 4x$ is a first-order linear differential equation in $z$. Solve by using an integrating factor. The integrating factor for $z' + \frac{z}{x} = 4$ is $\text{IF} = \text{e}^{\textstyle \int \frac{1}{x} \text{d}x} = x$.

Apply Integrating Factor

Multiplying both sides by the integrating factor $x$, we get: $x z' + z = 4x$ becomes $x z' + z = 4x$. This can be written as $ \frac{\text{d}}{\text{d}x} (xz) = 4x$.

Integrate Both Sides

Integrate both sides: $\frac{\reit}{\reit x} (xz) = \reit \int 4x \reit x$. This results in $xz = 2x^2 + C$ solving for $z$, yields: $z = 2x + \frac{C}{x} $.

Reverse Substitution

Reverse the substitution $z = y'$, thus $y' = 2x + \frac{C}{x}$. Integrate again to find $y$: $y = x^2 + C \reit \reit (\frac{1}{x}) x$.

Final Solution

After integration and including the constant of integration: $y = x^2 + C \reit (\textstyle \frac{x}) D$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

differential equations

Differential equations are mathematical equations that involve functions and their derivatives. They describe how a particular quantity changes over time or space. For example, when we look at the motion of a car, we can use a differential equation to model the car's velocity and its change over time.

integration

Integration is a fundamental concept in calculus that involves finding the antiderivative of a function. It is the inverse operation of differentiation. In the context of differential equations, integration is often used to find a function given its derivative. For instance, if we know that the derivative of a function is 2x, we can use integration to determine that the original function is x^2 + C, where C is a constant.

linear differential equations

Linear differential equations are equations where the dependent variable and its derivatives appear linearly (i.e., they are not multiplied together or raised to any power other than one). These equations can usually be written in the form: \[ a_n (x) y^{(n)}(x) + a_{n-1}(x) y^{(n-1)}(x) + \text{...} + a_1(x) y^{\treble}(x) + a_0(x) y(x) = g(x) \] In this equation, the coefficients $ a_n(x), a_{n-1}(x), \etc. $ are functions of the independent variable x, and $ g(x) $ is the non-homogeneous part of the equation. Solving these equations often requires finding an integrating factor or using methods like variation of parameters.

second order derivatives

Second order derivatives represent the rate of change of the rate of change of a function. In other words, it provides insight into the curvature or concavity of the function. In the equation, $ y'' = f(x, y, y') $, $ y'' $ is the second derivative of $ y $. For example, in the motion of an object, the first derivative of position with respect to time is velocity, and the second derivative is acceleration. Solving second order differential equations often involves two steps: converting the higher-order equation into a system of lower-order equations, and then solving those equations using methods like integration.

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