Find the general solution of the second-order differential equation $3y^{″}+2y^{\prime }+y=0$.

• A. $y=C_1{x}^2+C_{2}x$
• B. $y=C_1{e}^{-\frac{x}{3}}+C_2{e}^{-x}$
• C. $y=C_1{e}^{\frac{x}{3}}+C_2{e}^x$
• D. $y=C_1\sin \left(x\right)+C_2\cos \left(x\right)$

Answer: (B)

Verified step by step guidance

1. Rewrite the given second-order differential equation in standard form: \( y'' + \frac{2}{3}y' + \frac{1}{3}y = 0 \). This is a linear homogeneous differential equation with constant coefficients.
2. Assume a solution of the form \( y = e^{rx} \), where \( r \) is a constant to be determined. Substitute \( y = e^{rx} \) into the differential equation to obtain the characteristic equation.
3. The characteristic equation is \( 3r^2 + 2r + 1 = 0 \). Solve this quadratic equation using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = 2, c = 1 \).
4. Calculate the discriminant \( \Delta = b^2 - 4ac \). If \( \Delta > 0 \), the roots are real and distinct. If \( \Delta = 0 \), the roots are real and repeated. If \( \Delta < 0 \), the roots are complex conjugates.
5. Based on the nature of the roots, write the general solution. For real and distinct roots, the solution is \( y = C_1 e^{r_1 x} + C_2 e^{r_2 x} \). For repeated roots, the solution is \( y = (C_1 + C_2x)e^{r_1 x} \). For complex conjugate roots \( r = \alpha \pm \beta i \), the solution is \( y = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \).