Question

find the general solution of the given differential equation. $$ 9 y^{\prime \prime}+9 y^{\prime}-4 y=0 $$

Short answer

Question: Determine the general solution of the given second-order linear homogeneous differential equation: $$9y'' + 9y' - 4y = 0$$ Answer: The general solution of the given differential equation is $$y(x) = C_1 e^{\frac{1}{3}x} + C_2 e^{-\frac{4}{3}x}$$ where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Obtain the Characteristic Equation

Write the characteristic equation corresponding to the given differential equation. The characteristic equation will have the form: $$ar^2 + br + c = 0$$ where a, b, and c are the coefficients of the differential equation. In this case, the characteristic equation is: $$9r^2 + 9r - 4 = 0$$

Solve the Characteristic Equation

Solve the quadratic equation $$9r^2 + 9r - 4 = 0$$ Using the quadratic formula, we get: $$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Substituting a=9, b=9, and c=-4, we obtain: $$r = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 9 \cdot (-4)}}{2 \cdot 9}$$ After calculation, we find the roots as: $$r_1 = \frac{1}{3},\ r_2 = -\frac{4}{3}$$

Construct the General Solution

Since the roots are real and distinct, the general solution of the given differential equation is: $$y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}$$ Substituting the roots, we get the general solution: $$y(x) = C_1 e^{\frac{1}{3}x} + C_2 e^{-\frac{4}{3}x}$$ where \(C_1\) and \(C_2\) are arbitrary constants.

Key concepts

Characteristic Equation

When dealing with a linear differential equation like the one in the exercise, the concept of a **Characteristic Equation** is crucial. This is a polynomial equation derived from the differential equation itself and helps in finding the nature of solutions.

• The original differential equation you are working with is possibly daunting, but breaking it into smaller parts, like obtaining the characteristic equation, simplifies the process.
• The characteristic equation has the standard quadratic form: \[ar^2 + br + c = 0\]where \(a\), \(b\), and \(c\) are constants derived directly from the given differential equation.
• This linkage is straightforward: the coefficient of \(y''\) becomes \(a\), the coefficient of \(y'\) becomes \(b\) and the coefficient of \(y\) comes from \(c\).

For example, in the exercise, the differential equation had the form:\[9 y^{\prime \prime}+9 y^{\prime}-4 y=0\]leading to the characteristic equation:\[9r^2 + 9r - 4 = 0\] which is a nice, clean quadratic form easy to manage once you identify your values for \(a\), \(b\), and \(c\).

Quadratic Formula

When you have a characteristic equation that is quadratic, like here, the **Quadratic Formula** is your go-to method for finding solutions, or roots, of that equation. It provides roots based on the coefficients you identified.

• The formula itself is given by:\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• This formula will give you two possible roots because of the "\(\pm\)" symbol.
• Plugging in the values from our characteristic equation, \(a = 9\), \(b = 9\), and \(c = -4\), we use the quadratic formula to find:\[r = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 9 \cdot (-4)}}{2 \cdot 9}\]
• This step practically opens the door to finding your final solution by determining the nature and values of the roots.

By solving these, you discover: \(r_1 = \frac{1}{3}\)and\(r_2 = -\frac{4}{3}\), which direct us toward constructing the general solution.

General Solution

With your roots in hand from the quadratic formula, you can construct the **General Solution** to the differential equation. This solution gives you a formula expressing all potential solutions your original equation could have, using your found roots.

• The roots determine the form of the general solution. For real and distinct roots, the solution takes the form:\[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\]
• Here, \(C_1\) and \(C_2\) are arbitrary constants, which allow for a whole family of functions fitting the differential equation.
• Substituting the specific roots obtained \(r_1 = \frac{1}{3}\) and \(r_2 = -\frac{4}{3}\), the general solution becomes:\[y(x) = C_1 e^{\frac{1}{3}x} + C_2 e^{-\frac{4}{3}x}\]

This compact representation ensures all possible values that satisfy the original differential equation are included.

Real and Distinct Roots

Identifying **Real and Distinct Roots** in the context of a characteristic equation informs us about how a differential equation behaves and how its solutions look.

• A quadratic equation’s discriminant \(b^2 - 4ac\) tells you about the nature of the roots.
• If the discriminant is positive, the roots are real and distinct. This means they are different numbers and the solution grows or decays involving distinct exponential terms.
• In our exercise, the discriminant was:\[9^2 - 4 \cdot 9 \cdot (-4) = 81 + 144 = 225\] Being positive, it confirmed real, distinct roots:
• This characteristic leads to a general solution formed by two different exponentials, involving terms like \(e^{r_1 x}\) and \(e^{r_2 x}\).

Real and distinct roots provide clarity on the behavior of functions derived from differential equations, helping predict trends exemplified by the specific growth rates embedded in these exponential terms.