Problem 24 Solve the following first-order ... [FREE SOLUTION]

Chapter 14: Problem 24

Solve the following first-order equations for the boundary conditions given: (a) \(y^{\prime}-(y / x)=1, \quad y(1)=-1\) (b) \(y^{\prime}-y \tan x=1, \quad y(\pi / 4)=3 ;\) (c) \(y^{\prime}-y^{2} / x^{2}=1 / 4, \quad y(1)=1 ;\) (d) \(y^{\prime}-y^{2} / x^{2}=1 / 4, \quad y(1)=1 / 2\).

Short Answer

Expert verified

For (a), the solution is \( y = x (\ln|x| - 1) \). For (b), \( y = \frac{1 + \sin x}{\cos x} \). For (c), \( y = \sqrt{\frac{x^2}{2} + \frac{1}{2}} \). For (d), \( y = \sqrt{\frac{x^2}{2} - \frac{1}{4}} \).

Step by step solution

- Identifying the Equation Type for (a)

The given equation is a linear first-order differential equation of the form: \[ y' - \frac{y}{x} = 1 \]Identify it as a separable differential equation.

- Solving the Homogeneous Part for (a)

Solve the homogeneous equation: \[ y' - \frac{y}{x} = 0 \]By separating variables, we get: \[ \frac{y'}{y} = \frac{1}{x} \]Integrate both sides: \[ \ln|y| = \ln|x| + C_1 \rightarrow y = C_2 x \]

- Using Variation of Parameters for (a)

Let the particular solution be: \[ y = u(x)x \]Substitute and differentiate the particular solution: \[ y' = u'x + u \Rightarrow u'x + u - \frac{ux}{x} = 1 \Rightarrow u'x = 1 \Rightarrow u = \ln|x| + C_3 \]Thus, the solution is: \[ y = x (\ln|x| + C_3) \]

- Applying Boundary Condition for (a)

Given: \[ y(1) = -1 \]Substitute to find \( C_3 \): \[ -1 = 1(\ln 1 + C_3) \Rightarrow C_3 = -1 \]So, the solution is: \[ y = x (\ln|x| - 1) \]

- Identifying and Solving Equation (b)

Given equation: \[ y' - y \tan x = 1 \]Solve the homogeneous part: \[ y' - y \tan x = 0 \Rightarrow \frac{y'}{y} = \tan x \Rightarrow \ln|y| = -\ln|\cos x| + C_4 \Rightarrow y = \frac{C_5}{\cos x} \]

- Finding Particular Solution for (b)

Using variation of parameters, let: \[ y = u(x) \frac{1}{\cos x} \Rightarrow y' = u' \frac{1}{\cos x} - u \frac{\sin x}{\cos^2 x} \]Substitute back into the original equation to find: \[ u' \frac{1}{\cos x} = 1 \]Integrate: \[ u = \sin x + C_6 \]Thus, the general solution is: \[ y = (\sin x + C_6) \frac{1}{\cos x} \]

- Applying Boundary Condition for (b)

Given: \[ y(\pi/4) = 3 \]Substitute to find \( C_6 \): \[ 3 = \frac{\sin(\pi/4) + C_6}{\cos(\pi/4)} \Rightarrow 3 = \frac{\frac{\sqrt{2}}{2} + C_6}{\frac{\sqrt{2}}{2}} \Rightarrow 3 = 1 + 2C_6 \Rightarrow C_6 = 1 \]So, the solution is: \[ y = \frac{1 + \sin x}{\cos x} \]

- Identifying the Nonlinear Equation for (c)

The given equation is: \[ y' - \frac{y^2}{x^2} = \frac{1}{4} \]Identify it as a Bernoulli differential equation.

- Linearizing the Equation for (c)

Transform it by letting: \[ v = y^2 \rightarrow v' = 2yy' \] Substitute and transform: \[ 2yy' - \frac{y^2}{x^2} = \frac{1}{4} \rightarrow \frac{dv}{dx} - \frac{v}{2x^2} = \frac{1}{4} \]

- Solving the Transformed Equation for (c)

The equation is now linear: \[ v' - \frac{v}{2x^2} = \frac{1}{8} \]Solve using integrating factor: \[ u(x) = \int \frac{1}{2x^2} dx = -\frac{1}{2x} \]The particular solution is: \[ v = \frac{x^2}{2} + C \rightarrow y = \sqrt{\frac{x^2}{2} + C} \]

- Applying Boundary Condition for (c)

Given: \[ y(1) = 1 \]Substitute to find C: \[ 1 = \sqrt{\frac{1}{2} + C} \rightarrow 1 = \sqrt{\frac{1}{2} + C} \rightarrow 1 = \sqrt{\frac{1}{2} + C} \rightarrow \frac{1}{2} + C = 1 \rightarrow C = \frac{1}{2} \]So, the solution is: \[ y = \sqrt{\frac{x^2}{2} + \frac{1}{2}} \]

- Identifying the Nonlinear Equation for (d)

The given equation is the same as in part (c): \[ y' - \frac{y^2}{x^2} = \frac{1}{4} \]Use the same method as part (c) to solve.

- Solving the Transformed Equation for (d)

We have: \[ 2yy' - \frac{y^2}{x^2} = \frac{1}{4} \rightarrow \frac{dv}{dx} - \frac{v}{2x^2} = \frac{1}{4} \] The solution is: \[ v = \frac{x^2}{2} + C \rightarrow y = \sqrt{\frac{x^2}{2} + C} \]

- Applying Boundary Condition for (d)

Given: \[ y(1) = \frac{1}{2} \]Substitute to find C: \[ \frac{1}{2} = \sqrt{\frac{1}{2} + C} \rightarrow \frac{1}{4} = \frac{1}{2} + C \rightarrow C = -\frac{1}{4} \]So, the solution is: \[ y = \sqrt{\frac{x^2}{2} - \frac{1}{4}} \]

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Key Concepts

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

Separable Differential Equations

First-order differential equations can often be solved by identifying them as separable differential equations. This type of differential equation can be written in the form: \[ \frac{dy}{dx} = g(x)h(y) \]To solve, follow these steps:

Rewrite the equation to separate variables \( dy \) on one side and \( dx \) on the other.
Integrate both sides to obtain \( y \) as a function of \( x \).
Develop the general solution, which includes an integration constant.

For example, in part (a) of the exercise, we start with \( y' - \frac{y}{x} = 0 \), which translates to \( y' = \frac{y}{x} \) and leads us to separate and integrate both sides: \( \int \frac{dy}{y} = \int \frac{dx}{x} \). Integrating gives us \( \ln|y| = \ln|x| + C_1 \), leading to the solution \( y = C_2 x \).

Variation of Parameters

Variation of Parameters is a method to find a particular solution to a non-homogeneous linear differential equation. Typically, this method involves assuming that a particular solution \( y_p \) can be expressed in terms of the homogeneous solution, but with parameters that vary with \( x \). The steps are:

Start with the homogeneous solution \( y_h \).
Assume a new form for \( y_p \), involving unknown functions.
Substitute \( y_p \) into the original differential equation to find these functions.
Solve the resulting equation for those functions and then combine with \( y_h \).

In our exercise, for part (a), we use this method assuming \( y = u(x)x \) and find that \( u'x = 1 \), leading us to \( u = \ln|x| + C_3 \). Therefore, the general solution is \( y = x (\ln|x| + C_3) \).

Boundary Conditions

Boundary conditions are essential to obtaining a specific solution from a general solution of a differential equation. These conditions provide specific values for the function or its derivatives at certain points. Implementing boundary conditions involves the following:

Substitute the given boundary values into the general solution.
Solve for any integration constants that appear in the general solution.
Express the final specific solution using the determined constants.

In part (a), the boundary condition is \( y(1) = -1 \). By substituting this into \( y = x (\ln|x| + C_3) \), we solve for \( C_3 \), obtaining \( C_3 = -1 \), leading to the specific solution \( y = x ( \ln|x| - 1 ) \).

Bernoulli Differential Equation

A Bernoulli differential equation is a special type of nonlinear differential equation of the form:\[ y' + P(x)y = Q(x)y^n \]This can be linearized by an appropriate substitution. The steps to solve a Bernoulli equation are:

Divide the entire equation by \( y^n \).
Introduce a new variable \( v = y^{1-n} \).
Transform the original equation into a linear differential equation in terms of \( v \).
Solve the linear differential equation for \( v \).
Transform back to the original variable \( y \).

In part (c), we start with \( y' - \frac{y^2}{x^2} = \frac{1}{4} \). Assuming \( v = y^2 \), we transform this into \( \frac{dv}{dx} - \frac{v}{2x^2} = \frac{1}{4} \), a linear form, which we solve to get \( v \). This ultimately gives the solution \( y = \sqrt{\frac{x^2}{2} + \frac{1}{2}} \).

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