Question

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

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: \[ y = \sqrt{\frac{x^2}{2} - \frac{1}{4}} \]

Key concepts

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 \( y = \sqrt{\frac{x^2}{2} + \frac{1}{2}} \).