Question

For the given differential equation, (a) Determine the complementary solution, \(y_{c}(t)=c_{1} y_{1}(t)+c_{2} y_{2}(t)\). (b) Use the method of variation of parameters to construct a particular solution. Then form the general solution. \(y^{\prime \prime}+4 t y^{\prime}+\left(2+4 t^{2}\right) y=t^{2} e^{-t^{\prime}}\). [The functions \(y_{1}(t)=e^{-t^{*}}\) and \(y_{2}(t)=t e^{-t^{t}}\) are both solutions of the homogeneous equation.]

Short answer

Question: Determine the general solution for the given non-homogeneous differential equation. \[y''+4ty'+(2+4t^{2})y=t^{2}e^{-t^{2}}\] Following the step by step solution provided, find the general solution. Answer: The general solution for the given non-homogeneous differential equation is: \[y(t) = c_{1}e^{-t^2} + c_{2}te^{-t^2} - \left(\int t e^{-t^{2}}\right)e^{-t^{2}} + t\left(\int e^{-t^{2}}\right)e^{-t^2}\] where \(c_1\) and \(c_2\) are constants determined by initial conditions or boundary conditions.

Step-by-step solution

Differential Equation in Standard Form

The given differential equation is: \[y^{\prime \prime}+4ty^{\prime}+(2+4t^{2})y=t^{2}e^{-t^{2}}.\]

Complementary Solution

Since the complementary functions \(y_1(t) = e^{-t^2}\) and \(y_2(t) = te^{-t^2}\) are already given as part of the problem, the complementary solution \(y_c(t)\) is a linear combination of these functions: \[y_{c}(t) = c_{1}e^{-t^2} + c_{2}te^{-t^2}.\]

Wronskian of Complementary Functions

We will first find the Wronskian \(W(t)\). It is the determinant of the matrix: \[W(t)=\begin{vmatrix} e^{-t^{2}} & t e^{-t^{2}} \\ -2te^{-t^{2}} & e^{-t^{2}}-2t^{2} e^{-t^{2}} \end{vmatrix}\] Calculating the determinant, we get: \[W(t)=e^{-t^{2}}\left( e^{-t^{2}}-2t^{2} e^{-t^{2}}\right)-(t e^{-t^{2}})\left( -2te^{-t^{2}}\right)\] \[W(t)=e^{-2t^{2}}-2t^{2} e^{-2t^{2}}+2t^2 e^{-2t^{2}}= e^{-2t^{2}}.\]

Applying the Variation of Parameters Method

Now we have to construct a particular solution using the variation of parameters method. Our particular solution will have the form: \[y_p(t) = u_1(t)y_1(t) + u_2(t)y_2(t)\] To find the functions, \(u_1(t)\) and \(u_2(t)\), we need to solve the following pair of equations: \[u_1'(t)y_1(t) + u_2'(t)y_2(t) = 0\] \[u_1'(t)y_1'(t) + u_2'(t)y_2'(t) = G(t)\] In our case, \(G(t) = t^2e^{-t^2}\). Therefore, we have the following pair of equations: \[u_1'(t)e^{-t^{2}}+u_2'(t)te^{-t^{2}}=0\] \[u_1'(-2te^{-t^{2}})+u_2'(e^{-t^{2}}-2t^{2}e^{-t^{2}})=t^{2}e^{-t^{2}}\] We integrate both equations to find \(u_1(t)\) and \(u_2(t)\): \[u_1(t) = -\int t e^{-t^{2}}\] \[u_2(t) = \int e^{-t^{2}}\]

Particular Solution

Using the variation of parameters method from Step 4, we have the particular solution \(y_p(t)\) as: \[y_p(t) = -\left(\int t e^{-t^{2}}\right)e^{-t^{2}} + t\left(\int e^{-t^{2}}\right)e^{-t^2}\]

General Solution

Finally, combining the complementary and particular solutions, we have the general solution of the given differential equation: \[y(t) = y_c(t) + y_p(t) = c_{1}e^{-t^2} + c_{2}te^{-t^2} - \left(\int t e^{-t^{2}}\right)e^{-t^{2}} + t\left(\int e^{-t^{2}}\right)e^{-t^2}\] where \(c_1\) and \(c_2\) are constants determined by initial conditions or boundary conditions.

Key concepts

Complementary Solution

In solving differential equations, the complementary solution is crucial as it represents the solution of the homogeneous equation associated with the original differential equation. Specifically, this solution is derived from a linear combination of the solutions of the homogeneous equation. For the given exercise, the homogeneous equation is obtained by setting the non-homogeneous part to zero. Here, the solutions,

• \(y_1(t) = e^{-t^2}\)
• \(y_2(t) = te^{-t^2}\)

are provided and are fundamental in forming the complementary solution. When combined linearly, they yield:\[y_{c}(t) = c_{1} e^{-t^2} + c_{2} t e^{-t^2}\]where \(c_1\) and \(c_2\) are arbitrary constants. This expression represents all possible solutions to the homogeneous equation and forms a foundation for further analysis.

Variation of Parameters

The variation of parameters is a powerful technique to find a specific solution (particular solution) of a non-homogeneous differential equation. This method is effective when the complementary solution is already known. The strategy involves constructing a particular solution of the form:

• \(y_p(t) = u_1(t)y_1(t) + u_2(t)y_2(t)\)

In this exercise, functions \(u_1(t)\) and \(u_2(t)\) are determined via a special system of equations derived from the differential equation and its complementary solution. Specifically:

• \(u_1'(t)y_1(t) + u_2'(t)y_2(t) = 0\)
• \(u_1'(t)y_1'(t) + u_2'(t)y_2'(t) = G(t)\)

Where \(G(t)\) is the non-homogeneous part, here being \(t^2 e^{-t^2}\). Solving these equations allows us to integrate and find \(u_1(t)\) and \(u_2(t)\), thus constructing the particular solution.

Particular Solution

The particular solution of a differential equation provides insight into the influence of external forces or inputs represented by the non-homogeneous component. In this exercise, the particular solution is achieved using the variation of parameters method. The expression for this solution is:\[y_p(t) = -\left(\int t e^{-t^{2}}\right)e^{-t^{2}} + t\left(\int e^{-t^{2}}\right)e^{-t^2}\]This complicated expression results from integrating the functions \(u_1(t)\) and \(u_2(t)\) derived in the variation of parameters process. The integrals involved here are generally complex and require a keen understanding of calculus to resolve completely. The particular solution accounts for the specific response to the given input in the form of \(t^2 e^{-t^2}\).

General Solution

The general solution of a differential equation comprises both the complementary solution and a particular solution. It embodies all possible solutions to the differential equation. In this exercise, the general solution is expressed as:\[y(t) = y_c(t) + y_p(t) \]Combining:

• \(y_{c}(t) = c_{1}e^{-t^2} + c_{2}te^{-t^2}\)
• \(y_p(t) = -\left(\int t e^{-t^{2}}\right)e^{-t^{2}} + t\left(\int e^{-t^{2}}\right)e^{-t^2}\)

results in:\[y(t) = c_{1}e^{-t^2} + c_{2}te^{-t^2} - \left(\int t e^{-t^{2}}\right)e^{-t^{2}} + t\left(\int e^{-t^{2}}\right)e^{-t^2}\]This comprehensive solution can be fitted to initial conditions to resolve for constants \(c_1\) and \(c_2\), customizing the final behavior of the system.

Wronskian

The Wronskian is a determinant used in the mathematical analysis of differential equations to examine the linear independence of solutions. In this exercise, the Wronskian is essential for confirming that the given solutions \(y_1(t)\) and \(y_2(t)\) are linearly independent, ensuring the validity of the variation of parameters approach.The Wronskian is computed using:\[W(t)=\begin{vmatrix} e^{-t^{2}} & t e^{-t^{2}} \ -2te^{-t^{2}} & e^{-t^{2}}-2t^{2} e^{-t^{2}} \end{vmatrix}\]Solving the determinant gives:\[W(t)=e^{-2t^{2}}\]Since \(W(t) eq 0\), the independence of the solutions is confirmed. This result is crucial for the subsequent steps in solving the differential equation using the variation of parameters, as it assures that \(y_1(t)\) and \(y_2(t)\) form a fundamental solution set.