Question

Find the general solution of the given differential equation. $$ y^{\prime \prime}+5 y^{\prime}=0 $$

Short answer

Question: Determine the general solution of the given differential equation: \(y''+9y=t^{2}e^{3t}+6\). Answer: The general solution of the given differential equation is \(y(t) = c_1 \cos(3t) + c_2 \sin(3t) - 2te^{3t} + 3t^2e^{3t} + \frac{2}{3}\), where \(c_1\) and \(c_2\) are constants.

Step-by-step solution

Find the homogeneous equation

The homogeneous equation corresponding to the given equation is: $$ y^{\prime \prime}+9y=0 $$

Solve the homogeneous equation

The characteristic equation for the homogeneous equation is: $$ r^2+9=0 $$ Solve for r: $$ r = \pm 3i $$ Now, we can write the homogeneous solution as: $$ y_h(t) = c_1 \cos(3t) + c_2 \sin(3t) $$ Where \(c_1\) and \(c_2\) are constants.

Determine the appropriate form for the particular solution

The non-homogeneous part of the given equation is: $$ t^2e^{3t} + 6 $$ The particular solution will have the form: $$ y_p(t) = Ate^{3t} + Bt^2e^{3t}+C $$ Where A, B, and C are constants to be determined.

Find the derivatives of the particular solution and substitute into the given equation

The first and second derivatives of the particular solution are: $$ y_p^{\prime}(t) = (3A+B)t^1e^{3t} + (3A+2B)t^2e^{3t} $$ and $$ y_p^{\prime \prime}(t) = (9A + 6B)te^{3t} + (9A+12B)t^2e^{3t} $$ Now, substitute \(y_p\), \(y_p^{\prime}\), and \(y_p^{\prime \prime}\) into the given equation: $$ (9A + 6B)te^{3t} + (9A+12B)t^2e^{3t} + 9(Ate^{3t} + Bt^2e^{3t}+C) = t^2e^{3t}+6 $$

Solve for A, B, and C

Comparing coefficients on both sides of the equation, we get the following system of equations: $$ \begin{cases} 9A + 6B = 0 \\ 9A + 12B = 1 \\ 9C = 6 \end{cases} $$ By solving this system, we obtain: A = -2, B = 3, and C = 2/3.

Write the particular solution using the determined constants

With A, B, and C found, the particular solution is: $$ y_p(t) = -2te^{3t} + 3t^2e^{3t} + \frac{2}{3} $$

Combine the homogeneous and particular solutions

Finally, the general solution is the sum of the homogeneous and particular solutions: $$ y(t) = y_h(t) + y_p(t) = c_1 \cos(3t) + c_2 \sin(3t) - 2te^{3t} + 3t^2e^{3t} + \frac{2}{3} $$

Key concepts

Homogeneous Equation

When we refer to a homogeneous equation in the context of differential equations, we're talking about an equation without any external forces or inputs. In other words, it's a setup where the function and its derivatives are set to zero. With the given differential equation,
\(y^{\' \'}+9y=t^{2}e^{3t}+6\), the homogeneous part is obtained by removing the non-zero terms on the right side, resulting in
\(y^{\' \'}+9y=0\).
Why does it matter? It matters because the solution to the homogeneous equation, often denoted as \(y_h(t)\), serves as a building block for finding the general solution to the entire differential equation. It represents the system's behavior in the absence of forcing terms such as \(t^{2}e^{3t}+6\) in our case. In practice, solving the homogeneous equation is often a crucial first step, as it enables us to understand the underlying structure before additional complexities are added.

Characteristic Equation

The characteristic equation is at the heart of solving linear homogeneous differential equations with constant coefficients. From our original homogeneous equation \(y^{\' \'} + 9y = 0\), we derive the characteristic equation by assuming a solution in the form of \(e^{rt}\), leading us to
\(r^2 + 9 = 0\).
Solving for \( r \), we find the roots to be \(r = \text{pm}3i\). These complex roots tell us that the solution to our differential equation will involve trigonometric functions corresponding to the imaginary part of the roots, namely \(\text{cos}(3t)\) and \(\text{sin}(3t)\).
Importance in the overall problem: Once we've identified the roots of the characteristic equation, we have a powerful direction for constructing the general solution. We use the real parts of the roots to determine the exponential component of the solution and the imaginary parts for the oscillatory (trigonometric) components. This process simplifies the potentially complex work of solving differential equations, giving us a clear formula to reach the solution.

Particular Solution

Moving beyond the general behavior captured by the homogeneous equation, the particular solution, \(y_p(t)\), accounts for the specific impacts of non-zero terms in the differential equation. For our example, where the non-homogeneous term is \(t^2e^{3t} + 6\), we anticipate a particular solution of the form
\(y_p(t) = Ate^{3t} + Bt^2e^{3t} + C\), with \(A\), \(B\), and \(C\) as constants to be determined.
What's crucial here is selecting a form for \(y_p(t)\) that won't overlap with the homogeneous solution, allowing for a clear distinction when both are combined for the general solution. The exercise guidance recommends ensuring the particular solution is logically structured and systematically solved for, a task often involving matching coefficients and utilizing algebra to find the constants. This detail-oriented step is what allows us to properly handle the external forces affecting the system's behavior, represented by the non-homogeneous part of the equation. The end goal is to have \(y_p(t)\) encapsulate the essence of the non-homogeneous factors. Ultimately, the general solution of the differential equation combines the predictable patterns from the homogeneous solution with the unique characteristics of the particular solution.