Question

Find the particular integral of each equation: (a) \(y^{\prime \prime}(t)-2 y^{\prime}(t)+5 y=2\) (b) \(y^{\prime \prime}(t)+y^{\prime}(t)=7\) \((c) y^{\prime \prime}(t)+3 y=9\) \((d) y^{\prime \prime}(t)+2 y^{\prime}(t)-y=-4\) \((e) y^{\prime \prime}(t)=12\)

Short answer

(a) \(y_p = \frac{2}{5}\), (b) \(y_p = 7t\), (c) \(y_p = 3\), (d) \(y_p = 4\), (e) \(y_p = 6t^2 \).

Step-by-step solution

Identify the Form of the Particular Integral

First, identify the non-homogeneous part of each equation. This helps determine the form of the particular solution. For all given equations, the non-homogeneous part is a constant number. This suggests that the particular integral (PI) for each case is a constant, say \( y_p = C \).

Solve for the Constant in (a)

For equation (a) \(y^{\prime\prime}(t)-2y^{\prime}(t)+5y=2\), substitute \(y_p = C\) into the equation:\( -2(0) + 5C = 2 \).This simplifies to \( 5C = 2 \), giving \( C = \frac{2}{5} \).Thus, the particular integral is \( y_p = \frac{2}{5} \).

Solve for the Constant in (b)

For equation (b) \(y^{\prime\prime}(t)+y^{\prime}(t) = 7 \), substitute \(y_p = C\) into the equation:\( 0 + 0 = 7 \).Since the left term adds to zero and can't equal 7, the form must be adjusted to account for derivatives not being zero due to constants. Set \( y_p = 7t \).Then, verify: \( y_p' = 7 \) and \( y_p'' = 0 \) satisfy the equation as \( 0 + 7 = 7 \).Thus, \( y_p = 7t \) is a valid particular integral.

Solve for the Constant in (c)

For equation (c) \(y^{\prime\prime}(t) + 3y = 9 \), substitute \(y_p = C \) into the equation:\( 0 + 3C = 9 \).This simplifies to \( 3C = 9 \), so \( C = 3 \).Thus, the particular integral is \( y_p = 3 \).

Solve for the Constant in (d)

For equation (d) \(y^{\prime\prime}(t) + 2y^{\prime}(t) - y = -4 \), substitute \(y_p = C \) into the equation:\( 0 + 0 - C = -4 \).This simplifies to \( -C = -4 \), giving \( C = 4 \).Thus, the particular integral is \( y_p = 4 \).

Solve for the Constant in (e)

For equation (e) \(y^{\prime\prime}(t) = 12 \), substitute \(y_p = C \) into the equation:\( 0 = 12 \) doesn't work with \(y_p = C\). Setting \( y_p = 12t^2 / 2 \), then \( y_p'' = 12 \) verifies the solution is correct.Thus, \( y_p = 6t^2 \) is a valid particular integral.

Key concepts

Particular Integral

The particular integral plays a crucial role in solving non-homogeneous differential equations. It corresponds to a specific solution that accounts for the non-zero part of the differential equation — in simple terms, it is a solution that fits the additional aspects brought in by the non-homogeneous part.

In the exercise examples, the non-homogeneous part was often a constant. Therefore, we initially tried a constant form for the particular integral, denoted as \( y_p = C \). The idea is to find a specific constant value that satisfies the differential equation. For instance, in equation (a), by substituting \( y_p = C \) into the differential equation and simplifying, we could determine \( C = \frac{2}{5} \).

This approach highlights the importance of understanding the form of your non-homogeneous function to decide the right form for your particular integral.

Non-Homogeneous Differential Equation

A non-homogeneous differential equation differs from a homogeneous one by having a term on its right-hand side that is not zero. This term, often referred to as the non-homogeneous part, could be a function of \( t \), constants, or more complex expressions.

The presence of this extra term means solutions aren't purely determined by their derivatives and require addressing this additional element in the equation.

• For a constant non-homogeneous term, a constant particular integral may be suitable, as seen in some of the exercise examples.
• For polynomial terms, polynomial forms for the particular integral are tried.

The main task is to adjust the approach to the particular integral according to what the non-homogeneous element demands.

Second-Order Differential Equations

Second-order differential equations are characterized by the presence of the second derivative of a function. They are commonly encountered in physical systems modeling and engineering problems.

The equations presented in the exercise are second-order, necessitating attention to both first and second derivatives when calculating solutions. They often take forms such as:

• \( y''(t) + ay'(t) + by = g(t) \)

where \( g(t) \) is the non-homogeneous part. The order of an equation tells us about the complexity and the kind of solution techniques required. Second-order equations typically require both a complementary function (derived from the homogeneous equation) and a particular integral for solution completion.

Solution Techniques for Differential Equations

Solving differential equations requires a strategic approach based on their type and form. Here, various techniques apply depending on whether the equation is homogeneous or non-homogeneous, and the nature of the non-homogeneous term if present.

For the given exercise, these techniques included:

• Identifying the non-homogeneous term to choose an appropriate form for the particular integral.
• Substituting the trial solution form back into the differential equation to solve for unknown constants or functions.

Each equation type demands its tactical method, from guessing potential forms of solutions to using systematic methods like undetermined coefficients or variation of parameters.

Building a solution often involves combining the particular integral with the complementary solution (which solves the homogeneous version of the equation) to form the general solution.