Question

In each of Problems 1 through 10 find the general solution of the given differential equation. \(4 y^{\prime \prime}+12 y^{\prime}+9 y=0\)

Short answer

Question: Find the general solution of the given second order, linear, homogeneous differential equation: \[4y'' + 12y' + 9y = 0\] Answer: The general solution of the given differential equation is \[y(t) = C_1 e^{-\frac{3}{2}t} + C_2 te^{-\frac{3}{2}t}\], where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Write down the given differential equation

The given differential equation is: \[4y'' + 12y' + 9y = 0\]

Assume a solution of the form \(y = e^{rt}\)

Let's assume a solution for the differential equation of the form: \[y(t) = e^{rt}\]

Substitute and find the auxiliary equation

Take the first and second derivatives of the assumed solution: \[y'(t) = r e^{rt} \quad \text{and} \quad y''(t) = r^2 e^{rt}\] Substitute these derivatives and the assumed solution into the given differential equation: \[4(r^2 e^{rt}) + 12(re^{rt}) + 9(e^{rt}) = 0\] Since \(e^{rt} \neq 0\), we can divide both sides by \(e^{rt}\) and obtain the auxiliary equation: \[4r^2 + 12r + 9 = 0\]

Solve the auxiliary equation

Now, we'll solve for r in the equation: \[4r^2 + 12r + 9 = 0\] This is a quadratic equation which can be factored as \((2r + 3)^2 = 0\). Therefore, we have a repeated root of \(r = -\frac{3}{2}\).

Write down the final general solution

Since we have a repeated root, the general solution is in the form: \[y(t) = C_1 e^{-\frac{3}{2}t} + C_2 te^{-\frac{3}{2}t}\] where \(C_1\) and \(C_2\) are arbitrary constants.

Key concepts

Auxiliary Equation

To solve second-order linear differential equations like the one given, we need to convert them into a polynomial equation known as the auxiliary equation. This step simplifies the differential equation and allows us to find the roots, which are crucial for determining the solution.

The concept of the auxiliary equation involves replacing the differential terms with nth powers of a variable, often written as \(r\). For our differential equation:\[4y'' + 12y' + 9y = 0\]we assume a solution of the form \(y = e^{rt}\). Taking the derivatives, the substituted terms become a polynomial:

• \(y'(t) = re^{rt}\)
• \(y''(t) = r^2e^{rt}\)

Substituting these into the differential equation, and factoring out the exponential term, we derive the auxiliary equation:\[4r^2 + 12r + 9 = 0\]This quadratic equation contains all the necessary information for finding solutions to the differential equation.

General Solution

The solution to a differential equation is greatly informed by the roots of its auxiliary equation.The form of these roots dictates the type of solutions we can expect.

The general solution combines solutions that arise from these roots. Here, by solving the auxiliary equation \(4r^2 + 12r + 9=0\), we find it factors as \((2r + 3)^2 = 0\). This means the root is repeated: \(r = -\frac{3}{2}\). For each distinct root, we typically associate an exponential function in the form \(e^{rt}\).

For repeated roots, unlike distinct ones, the solution changes slightly:

• For a single root \(r_1\), the solutions are \(C_1 e^{r_1 t}\) and \(C_2 te^{r_1 t}\).

Thus, in our case, the general solution becomes:\[ y(t) = C_1 e^{-\frac{3}{2}t} + C_2 te^{-\frac{3}{2}t} \]where \(C_1\) and \(C_2\) are arbitrary constants, accounting for the homogeneity of the equation.

Repeated Root

In differential equations, a repeated root scenario occurs when a quadratic equation, like our auxiliary equation \(4r^2 + 12r + 9 = 0\), does not have distinct solutions.

Instead, the discriminant \(b^2 - 4ac\) might be zero, indicating that the quadratic can be expressed as a squared factor, resulting in a repeated root situation. Here, our equation factors as \((2r + 3)^2 = 0\) giving us the root \(r = -\frac{3}{2}\) with multiplicity two.

In comparison to distinct roots, repeated roots require a little twist in the solution: while a distinct root would simply give \(C e^{rt}\), a repeated root means adding an extra linear factor of \(t\) to deal with the multiplicity:

• This method prevents overlapping terms that occur with identical exponential factors.
• The sequence \(C_2 te^{r t}\) ensures we derive two linearly independent solutions.

This alteration is crucial for ensuring that our solution space has enough functions to encompass the general behavior of the differential equation.