Question

Use the method outlined in Problem 28 to solve the given differential equation. $$ t^{2} y^{\prime \prime}-2 t y^{\prime}+2 y=4 t^{2}, \quad t>0 ; \quad y_{1}(t)=t $$

Short answer

Answer: The general solution for the given differential equation is \(y(t) = c_1t -\frac{c_2}{2t^2} - 2t^2 + \frac{t^4}{4}\)

Step-by-step solution

Identify the complementary function

The given differential equation is $$t^2 y'' - 2ty' + 2y = 4t^2$$ with the first complementary function \(y_1(t) = t\). Since we have one complementary function, we can find the Wronskian \(W(t)\) which is given by $$W(t) = y_1(t)y_1'(t) = t(1) = t$$ Now, let's calculate the particular integral.

Find the particular integral

To find the particular integral \(y_p(t)\), we use variation of parameters method as follows: First, we find the second complementary function \(y_2(t)\) using the formula: $$y_2(t)=y_1(t)\int{\frac{1}{y_1^2(t)W(t)} \ dt}$$ Substituting the values, we get $$y_2(t)=t\int{\frac{1}{t^2 \cdot t} \ dt} = t\int{\frac{1}{t^3} \ dt}$$ Integrating with respect to \(t\), we get $$y_2(t)= -\frac{1}{2t^2}$$ Second, we find the particular integral \(y_p(t)\) using the formula: $$y_p(t)= -y_1(t)\int{y_2(t) Q(t) \ dt}+y_2(t)\int{y_1(t) Q(t) \ dt}$$ Here, \(Q(t)\) is the right-hand side of the given non-homogeneous differential equation, which is \(4t^2\). Substituting the values, we get $$y_p(t)=-t\int{(-\frac{1}{2t^2})(4t^2) \ dt}+(-\frac{1}{2t^2})\int{t(4t^2) \ dt}$$ Simplifying and integrating, we get $$y_p(t)=-t\int{2 \ dt}+(-\frac{1}{2t^2})\int{4t^3 \ dt}$$ $$y_p(t)=-2t^2+\frac{t^4}{4}$$

Obtain the general solution

Now that we have the complementary function and the particular integral, we can obtain the general solution for the given differential equation as follows: $$y(t) = c_1y_1(t) + c_2y_2(t) + y_p(t)$$ Substituting the values, $$y(t) = c_1t + c_2(-\frac{1}{2t^2}) + (-2t^2 + \frac{t^4}{4})$$ Thus, the general solution for the given differential equation is $$y(t) = c_1t -\frac{c_2}{2t^2} - 2t^2 + \frac{t^4}{4}$$

Key concepts

Complementary Function

The complementary function is a crucial part of solving differential equations. It represents the solution to the associated homogeneous equation, which in this case is \[ t^2 y'' - 2ty' + 2y = 0 .\]For such an equation, the solution space is formed by functions that satisfy the equation without any external input or forcing functions.

In this problem, one has already been provided: the complementary function is \(y_1(t) = t\). This is part of what we need to construct the solution set of the homogeneous equation.

The next step involves finding any additional linearly independent solutions through methods such as the Wronskian, to determine the full complementary function. The complementary function forms the "backbone" of the differential equation's solution, onto which we add particular solutions when dealing with non-homogeneous equations.

Variation of Parameters

Variation of parameters is a powerful method used to find particular solutions to non-homogeneous differential equations. It is especially useful when the differential equation is complex, and simple guessing methods for particular solutions are ineffective. The technique leverages already known solutions of the homogeneous differential equation, the complementary functions.

In our exercise, we apply variation of parameters by finding a second complementary function \(y_2(t)\) and using it together with \(y_1(t)\) to find the particular integral. The method uses integrals of combinations of these functions and the non-homogeneous part \(Q(t)\).

• Determine \(y_2(t)\) using \[y_2(t) = y_1(t) \int \frac{1}{y_1^2(t) W(t)} \, dt \]
• Use both \(y_1(t)\) and \(y_2(t)\) to form \(y_p(t)\), the particular integral.

Through these steps, variation of parameters grant us the ability to customize the solution to incorporate the non-homogeneous component \(Q(t) = 4t^2\).

Particular Integral

The particular integral, denoted as \(y_p(t)\), represents the specific solution to the non-homogeneous differential equation that accounts for the external influence represented by the non-zero right-hand side, \(4t^2\) in this case.

To find this, we use the method of variation of parameters, which involves:

• Using \(-y_1(t) \int y_2(t) Q(t) \, dt + y_2(t) \int y_1(t) Q(t) \, dt\) as the formula.
• Insert known functions \(y_1(t)\) and \(y_2(t)\) and compute the integrals, adjusting for \(Q(t) = 4t^2\).

The particular integral allows us to capture the effects of external forces on the system, effectively modifying the solution beyond what's dictated by the homogeneous equation.

General Solution

The general solution of a differential equation combines all possible solutions into a single expression. It is constructed by adding together:

• Complementary Function, \(c_1y_1(t) + c_2y_2(t)\).
• Particular Integral, \(y_p(t)\).

This approach provides the most comprehensive solution set for the differential equation.

In the exercise overhead, we compile the general solution for\[ t^2 y'' - 2ty' + 2y = 4t^2 \] as:\[ y(t) = c_1t - \frac{c_2}{2t^2} - 2t^2 + \frac{t^4}{4} .\]Here, the constants \(c_1\) and \(c_2\) are determined by boundary or initial conditions related to a specific situation. The general solution captures all diverse responses possible under different scenarios of initial or boundary conditions.