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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

01

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 \(g(x) = 4x\).
02

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: \(x y^{\riet^2}+ y^{\reit} = 4x\).
03

Reducing Order

To solve this, first reduce the order by substituting \(y' = z\) which implies \(y'' = z'\). Therefore, the equation becomes \(x z' + z = 4x\).
04

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 \(\text{IF} = \text{e}^{\textstyle \int \frac{1}{x} \text{d}x} = x\).
05

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 \( \frac{\text{d}}{\text{d}x} (xz) = 4x\).
06

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: \(z = 2x + \frac{C}{x} \).
07

Reverse Substitution

Reverse the substitution \(z = y'\), thus \(y' = 2x + \frac{C}{x}\). Integrate again to find \(y\): \(y = x^2 + C \reit \reit (\frac{1}{x}) x\).
08

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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Most popular questions from this chapter

A mass \(m\) falls under gravity (force \(m g\) ) through a liquid whose viscosity is decreasing so that the retarding force is \(-2 m v /(1+t),\) where \(v\) is the speed of \(m .\) If the mass starts from rest, find its speed, its acceleration, and how far it has fallen (in terms of \(g\) ) when \(t=1\).

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}+4 y^{\prime}+5 y=2 e^{-2 x} \cos x$$

An object of mass \(m\) falls from rest under gravity subject to an air resistance proportional to its speed. Taking the \(y\) axis as positive down, show that the differential equation of motion is \(m(d v / d t)=m g-k v,\) where \(k\) is a positive constant. Find \(v\) as a function of \(t,\) and find the limiting value of \(v\) as \(t\) tends to infinity; this limit is called the terminal speed. Can you find the terminal speed directly from the differential equation without solving it? Hint: What is \(d v / d t\) after \(v\) has reached an essentially constant value? Consider the following specific examples of this problem. (a) A person drops from an airplane with a parachute. Find a reasonable value of \(k\) (b) In the Millikan oil drop experiment to measure the charge of an electron, tiny electrically charged drops of oil fall through air under gravity or rise under the combination of gravity and an electric field. Measurements can be made only after they have reached terminal speed. Find a formula for the time required for a drop starting at rest to reach 99\% of its terminal speed.

The differential equation for the path of a planet around the sun (or any object in an inverse square force field) is, in polar coordinates, \(\frac{1}{r^{2}} \frac{d}{d \theta}\left(\frac{1}{r^{2}} \frac{d r}{d \theta}\right)-\frac{1}{r^{3}}=-\frac{k}{r^{2}}\). Make the substitution \(u=1 / r\) and solve the equation to show that the path is a conic section.

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