Question

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ (1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, \quad 0

Short answer

Given the functions \(y_1(t)=e^t\) and \(y_2(t)=t\), we first found their first and second derivatives. Then, we verified that \(y_1\) and \(y_2\) satisfy the homogeneous equation \((1-t)y^{\prime \prime} + ty^{\prime} - y = 0\). Next, we proposed a particular solution of the form \(y_p(t)=A(t-1)^{2}e^{-t}\) for the nonhomogeneous equation \((1-t)y^{\prime \prime} + ty^{\prime} - y = 2(t-1)^{2} e^{-t}\), found its first and second derivatives, and determined the constant A. Finally, we found the general solution to the problem: \(y(t) = c_1e^t + c_2t+ (t-1)^{2}e^{-t}\), where \(c_1\) and \(c_2\) are arbitrary constants.

Step-by-step solution

Finding the first and second derivatives of \(y_1\) and \(y_2\)

For \(y_1(t)=e^{t}\), - Find the first derivative: \(y_1^{\prime}(t)=\frac{d}{dt}\left(e^t\right)=e^t\) - Find the second derivative: \(y_1^{\prime \prime}(t)=\frac{d^2}{dt^2}\left(e^t\right)=e^t\) For \(y_2(t)=t\), - Find the first derivative: \(y_2^{\prime}(t)=\frac{d}{dt}\left(t\right)=1\) - Find the second derivative: \(y_2^{\prime \prime}(t)=\frac{d^2}{dt^2}\left(t\right)=0\)

Verify if \(y_1\) and \(y_2\) satisfy the homogeneous equation

The homogeneous equation is: \((1-t)y^{\prime \prime} + ty^{\prime} - y = 0\) For \(y_1\): \((1-t)y_1^{\prime \prime} + ty_1^{\prime} - y_1 = (1-t)(e^t) + t(e^t) - e^t = 0\), so \(y_1\) satisfies the homogeneous equation. For \(y_2\): \((1-t)y_2^{\prime \prime} + ty_2^{\prime} - y_2 = (1-t)(0) + t(1) - t = 0\), so \(y_2\) satisfies the homogeneous equation.

Finding a particular solution for the nonhomogeneous equation

The nonhomogeneous equation is: \((1-t)y^{\prime \prime} + ty^{\prime} - y = 2(t-1)^{2} e^{-t}\) Using the method of undetermined coefficients, we propose a particular solution of the form: \(y_p(t)=A(t-1)^{2}e^{-t}\), where A is a constant to be determined. Now we find the first and second derivatives of \(y_p\): - \(y_p^{\prime}(t) = A e^{-t} \left(2(t-1) - (t-1)^{2}\right)\) - \(y_p^{\prime \prime}(t) = A e^{-t} \left(-2 + 4(t-1) - (t-1)^{2}\right)\)

Substitute \(y_p\) and its derivatives into the nonhomogeneous equation

Now, we substitute \(y_p\) and its derivatives into the nonhomogeneous equation: \((1-t)(-A e^{-t} \left(-2 + 4(t-1) - (t-1)^{2}\right)) + t A e^{-t} \left(2(t-1) - (t-1)^{2}\right) - A(t-1)^{2}e^{-t} = 2(t-1)^{2} e^{-t}\) Simplifying the equation by canceling out the \(e^{-t}\) terms and collecting like terms, we get: \(A(4 - 4t + 2t^2) = 2t^2 - 4t + 2\) Comparing the coefficients of the terms, we find \(A=1\). So the particular solution is: \(y_p(t)=(t-1)^{2}e^{-t}\) Now we found the particular solution, and we already know \(y_1(t)\) and \(y_2(t)\) satisfy the homogeneous equation. So the general solution to the problem is: \(y(t) = c_1y_1(t) + c_2y_2(t) + y_p(t) = c_1e^t + c_2t+ (t-1)^{2}e^{-t}\), where \(c_1\) and \(c_2\) are arbitrary constants.

Key concepts

Homogeneous Equation

A homogeneous equation is a type of differential equation that, when set equal to zero, only involves the function and its derivatives but not external terms or functions. In simple terms, it does not include any 'outside' elements or 'source' terms in its expression.

For instance, an expression like \( (1-t)y^{\prime \prime} + ty^{\text{'}} - y = 0 \) is considered homogeneous because it lacks a specific function of \( t \) on the right-hand side of the equation. Problems involving homogeneous equations often require verification that certain functions are solutions, an action similar to 'plugging in' the proposed functions and their derivatives into the original equation to ensure the equality holds true.

In the given exercise, both functions \( y_1(t) = e^t \) and \( y_2(t) = t \) are tested to confirm their validity as solutions to the corresponding homogeneous equation.

Nonhomogeneous Equation

In contrast to homogeneous equations, a nonhomogeneous equation features an extra term that acts as an external 'forcing' function and is not a trivial zero. This additional component is often due to external influences on the system described by the equation.

The general form of the nonhomogeneous equation looks like \( a(t)y^{\prime \prime} + b(t)y^{\text{'}} + c(t)y = g(t) \) where \( g(t) \) is the nonhomogeneous or 'forcing' term. Essentially, nonhomogeneous equations describe a wider array of problems because of this ability to incorporate non-zero external terms into their structure.

In the provided exercise, the nonhomogeneous term is \( 2(t-1)^{2} e^{-t} \) and is the key difference from the homogeneous form. Our goal with such equations is to find the particular solution that satisfies this specific setup.

Method of Undetermined Coefficients

The method of undetermined coefficients is an approach to finding a particular solution to nonhomogeneous differential equations. The idea is to assume a form for the solution that has some 'adjustable' constants, which are then determined through substitution and simplification.

Typically, the proposed solution form will mirror the structure of the nonhomogeneous term. For example, if the nonhomogeneous term is a polynomial, the proposal for a particular solution will also be a polynomial, with arbitrary coefficients replacing the defined numerical coefficients.

In practice, it involves a few key steps:

• Guessing a form for the particular solution with undetermined coefficients.
• Differentiating this guessed form, often multiple times depending on the degree of the differential equation.
• Substituting this form and its derivatives back into the original nonhomogeneous equation.
• Solving the resulting algebraic equations to determine the values of the coefficients.

In our exercise, the method is used to deduce that the particular solution takes on the form \( y_p(t)=(t-1)^{2}e^{-t} \) by finding the coefficient \( A \) that satisfies the nonhomogeneous equation.

Particular Solution

A particular solution to a nonhomogeneous differential equation is a single, specific solution that satisfies not only the differential equation itself but also incorporates the nonhomogeneous term. It is one special 'instance' of a solution out of potentially many.

The particular solution is found after considering the homogeneous solutions to the equation. In the strategy of finding a complete solution to a nonhomogeneous differential equation, the particular solution is combined with the complementary solution (comprised of the homogeneous solutions) to form the general solution.

In our exercise, the particular solution \( y_p(t)=(t-1)^{2}e^{-t} \) was determined using the method of undetermined coefficients and uniquely corresponds to the nonhomogeneous part of the given equation. When we combine this with the complementary function, we obtain the full solution to the differential equation, which reveals how the system behaves in all possible scenarios encapsulated by the equation.