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.