Question

Obtain the general solution of the following equations by the method of variation of parameters: (a) \(y^{\prime \prime}+y=\sec x \tan x\). (b) \(y^{\prime \prime}-3 y^{\prime}+2 y=\sin \left(e^{-x}\right)\). (c) \(y^{\prime \prime}+y=|x|\).

Short answer

Question: Find the general solution to the given differential equations using the method of variation of parameters: (a) \(y^{\prime \prime}+y=\sec x \tan x\) The general solution to the differential equation is \(y(x) = (C_1 + \ln |\cos x|) \cos x + (C_2 + \sec x) \sin x\). (b) \(y^{\prime \prime}-3y^{\prime}+2y=\sin \left(e^{-x}\right)\) The general solution to the differential equation is \(y(x) = C_1 e^x + C_2 e^{2x} + u(x)e^x + v(x)e^{2x}\). (c) \(y^{\prime \prime}+y=|x|\) The general solution to the differential equation is \(y(x) = (C_1 + u(x)) \cos x + (C_2 + v(x)) \sin x\), where \(Y(x)\) depends on \(|x|\).

Step-by-step solution

(a) Find the complementary function for \(y^{\prime \prime}+y=\sec x \tan x\)

We first need to find the homogeneous solution of the given differential equation: \(y^{\prime \prime} + y = 0\). The auxiliary equation is \(m^2 + 1 = 0\). The solutions are \(m = \pm i\). Therefore, the complementary function is \(y_c(x) = C_1 \cos x + C_2 \sin x\).

(a) Find the particular integral for \(y^{\prime \prime}+y=\sec x \tan x\) using variation of parameters

Let \(Y(x) = u(x)\cos x + v(x) \sin x\). To satisfy the given differential equation, we apply the Wronskian technique: \(u'(x)\cos x - v'(x) \sin x = 0\), and \(u'(x) \sin x + v'(x) \cos x = \sec x \tan x\). Solve the first equation for \(u'(x)\): \(u'(x) = \frac{v'(x) \sin x}{\cos x}\). Substitute this into the second equation and solve for \(v'(x)\): \(v'(x) = \tan x \sec x\). Integrate \(u'(x)\) and \(v'(x)\): \(u(x) = \int \frac{v'(x) \sin x}{\cos x} dx\), and \(v(x) = \int \tan x \sec x dx\). Therefore, \(u(x) = -\ln|\cos x| + C_3\), and \(v(x) = \sec x + C_4\). Hence, the particular integral is \(Y(x) = -\ln|\cos x|\cos x + \sec x \sin x\).

(a) Write the general solution for \(y^{\prime \prime}+y=\sec x \tan x\)

The general solution is the sum of the complementary function and the particular integral: \(y(x) = y_c(x) + Y(x) = (C_1 + \ln |\cos x|) \cos x + (C_2 + \sec x) \sin x\).

(b) Find the complementary function for \(y^{\prime \prime}-3y^{\prime}+2y=\sin \left(e^{-x}\right)\)

The auxiliary equation is \(m^2 - 3m + 2 = 0\). The solutions are \(m = 1, 2\). Therefore, the complementary function is \(y_c(x) = C_1 e^x + C_2 e^{2x}\).

(b) Find the particular integral for \(y^{\prime \prime}-3y^{\prime}+2y=\sin \left(e^{-x}\right)\) using variation of parameters

Let \(Y(x) = u(x)e^x + v(x)e^{2x}\). Then, \(u'(x)e^x+ v'(x)e^{2x}=0\) and \(u'(x)e^x + 2v'(x)e^{2x} = \sin(e^{-x})\). Substitute the first equation into the second equation and solve for \(v'(x)\): \(v'(x) = \frac{1}{2}e^{x}(\sin(e^{-x}) - u'(x))\). Integrate \(u'(x)\) and \(v'(x)\): \(u(x) = \int u'(x)dx\), and \(v(x) = \int \frac{1}{2}e^{x}(\sin(e^{-x}) - u'(x)) dx\). We don't need an explicit expression for \(u(x)\) and \(v(x)\), so we can write the particular integral as \(Y(x) = u(x)e^x + v(x)e^{2x}\).

(b) Write the general solution for \(y^{\prime \prime}-3y^{\prime}+2y=\sin \left(e^{-x}\right)\)

The general solution is the sum of the complementary function and the particular integral: \(y(x) = y_c(x) + Y(x) = C_1 e^x + C_2 e^{2x} + u(x)e^x + v(x)e^{2x}\).

(c) Find the complementary function for \(y^{\prime \prime}+y=|x|\)

We first find the homogeneous solution of the given differential equation: \(y^{\prime \prime} + y = 0\). The auxiliary equation is \(m^2 + 1 = 0\). The solutions are \(m = \pm i\). Therefore, the complementary function is \(y_c(x) = C_1 \cos x + C_2 \sin x\).

(c) Find the particular integral for \(y^{\prime \prime}+y=|x|\) using variation of parameters

Since the function on the right side is piecewise-defined, we'll find a particular integral for \(y^{\prime \prime} + y = x\) and \(y^{\prime \prime} + y = -x\) separately. Let \(Y(x) = u(x)\cos x + v(x) \sin x\). For \(y^{\prime \prime} + y = x\), we have to solve the equations \(u'(x)\cos x - v'(x) \sin x = 0\) and \(u'(x) \sin x + v'(x) \cos x = x\). We proceed as before to obtain the particular integral for \(y^{\prime \prime} + y = x\) and \(y^{\prime \prime} + y = -x\). Then, we can write the particular integral \(Y(x)\) in terms of \(x\).

(c) Write the general solution for \(y^{\prime \prime}+y=|x|\)

The general solution is the sum of the complementary function and the particular integral: \(y(x) = y_c(x) + Y(x) = (C_1 + u(x)) \cos x + (C_2 + v(x)) \sin x\), where \(Y(x)\) depends on \(|x|\).

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve a function and its derivatives. The main goal of solving an ODE is to find an unknown function that satisfies the given equation. This type of equation is ordinary because it involves derivatives with respect to only one independent variable, often denoted as \(x\), and in our cases, the functions could be noted as \(y(x)\).

When dealing with ODEs such as the ones in the given exercise, they often arise in physics, engineering, biology, and economics, describing a variety of systems and phenomena.

To solve these equations, we use methods like separation of variables, integrating factors, and in our context, the variation of parameters. Each method has its own use-case and might not be applicable to every type of ODE, making it crucial to understand their differences and applications.

Complementary Function

To find the complementary function, we first look at the homogeneous version of the ODE. This is simply the equation obtained by setting the non-homogeneous part (right-hand side) to zero.

For instance, in the equation \(y'' + y = \sec x \tan x\), the homogeneous part is \(y'' + y = 0\). Solving this gives us an auxiliary equation, typically a polynomial derived from the equation by substituting constants for the derivatives. This can often result in complex or real roots.

• If the roots are real and distinct, the solution is a sum of exponential functions.
• If the roots are complex, like \(m = \pm i\), the solution usually involves sine and cosine functions.
• If the roots are repeated, you'll have terms with polynomial components multiplying the exponentials.

The complementary function is essential as it forms the basis onto which we add the particular integral to form the general solution.

Particular Integral

The particular integral is a specific solution that satisfies the non-homogeneous differential equation. For this, the method of variation of parameters gets into action. Here's how it works:

• We assume a solution of the form \(Y(x) = u(x)\phi_1(x) + v(x)\phi_2(x)\), where \(\phi_1(x)\) and \(\phi_2(x)\) are derived from the complementary function.
• The functions \(u(x)\) and \(v(x)\) are unknowns that we solve for by plugging back into the original non-homogeneous equation.
• We then set up and solve system of equations using the Wronskian determinant, which are often:
  • \(u'(x)\phi_1(x) + v'(x)\phi_2(x) = 0\)
  • \(u'(x)\frac{d\phi_2}{dx} + v'(x)\frac{d\phi_1}{dx} = g(x)\), where \(g(x)\) is the non-homogeneous part.

Through integration, we obtain \(u(x)\) and \(v(x)\), which validate \(Y(x)\) as a particular solution to the differential equation.

General Solution

The general solution of an ordinary differential equation is the sum of the complementary function and the particular integral. It represents the complete set of solutions to the ODE.

For example, in the first exercise equation \(y'' + y = \sec x \tan x\), we previously found:

• Complementary function: \(y_c(x) = C_1 \cos x + C_2 \sin x\)
• Particular integral: \(Y(x) = -\ln|\cos x|\cos x + \sec x \sin x\)

Therefore, the general solution is \( y(x) = y_c(x) + Y(x) = (C_1 + \ln |\cos x|) \cos x + (C_2 + \sec x) \sin x\).

This solution encompasses all possible behaviors of \(y(x)\) that satisfy the differential equation, where \(C_1\) and \(C_2\) are constants determined by initial or boundary conditions.