\( 2 y ^ { \prime \prime \prime } + 3 y ^ { \prime \prime } + 6 y ^ { \prime } - 4 y = 0 \)

To solve the differential equation \[ 2 y^{\prime\prime\prime} + 3 y^{\prime\prime} + 6 y^{\prime} - 4 y = 0, \] we start by finding the characteristic equation associated with this linear homogeneous differential equation. We assume a solution of the form \( y = e^{rt} \), where \( r \) is a constant. Substituting \( y = e^{rt} \) into the differential equation, we compute the derivatives: \[ y' = r e^{rt}, \quad y'' = r^2 e^{rt}, \quad y''' = r^3 e^{rt}. \] Substituting these into the differential equation gives: \[ 2 (r^3 e^{rt}) + 3 (r^2 e^{rt}) + 6 (r e^{rt}) - 4 (e^{rt}) = 0. \] Factoring out \( e^{rt} \) (which is never zero), we obtain the characteristic equation: \[ 2r^3 + 3r^2 + 6r - 4 = 0. \] Next, we need to solve the cubic equation \( 2r^3 + 3r^2 + 6r - 4 = 0 \). We can use the Rational Root Theorem to test for possible rational roots. Testing \( r = 1 \): \[ 2(1)^3 + 3(1)^2 + 6(1) - 4 = 2 + 3 + 6 - 4 = 7 \quad (\text{not a root}), \] Testing \( r = -1 \): \[ 2(-1)^3 + 3(-1)^2 + 6(-1) - 4 = -2 + 3 - 6 - 4 = -9 \quad (\text{not a root}), \] Testing \( r = 2 \): \[ 2(2)^3 + 3(2)^2 + 6(2) - 4 = 16 + 12 + 12 - 4 = 36 \quad (\text{not a root}), \] Testing \( r = -2 \): \[ 2(-2)^3 + 3(-2)^2 + 6(-2) - 4 = -16 + 12 - 12 - 4 = -20 \quad (\text{not a root}), \] Testing \( r = \frac{1}{2} \): \[ 2\left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) - 4 = 2\left(\frac{1}{8}\right) + 3\left(\frac{1}{4}\right) + 3 - 4 = \frac{1}{4} + \frac{3}{4} + 3 - 4 = 0 \quad (\text{is a root}), \] Now that we have found \( r = \frac{1}{2} \) as a root, we can perform polynomial long division to factor \( 2r^3 + 3r^2 + 6r - 4 \) by \( r - \frac{1}{2} \). Dividing \( 2r^3 + 3r^2 + 6r - 4 \) by \( r - \frac{1}{2} \): 1. Divide the leading term: \( 2r^3 \div r = 2r^2 \). 2. Multiply \( 2r^2 \) by \( r - \frac{1}{2} \): \( 2r^3 - r \). 3. Subtract: \( (2r^3 + 3r^2 + 6r - 4) - (2r^3 - r) = 3r^2 + 7r - 4 \). 4. Divide the leading term: \( 3r^2 \div r = 3r \). 5. Multiply \( 3r \) by \( r - \frac{1}{2} \): \( 3r^2 - \frac{3}{2} \). 6. Subtract: \( (3r^2 + 7r - 4) - (3r^2 - \frac{3}{2}) = 7r - 4 + \frac{3}{2} = 7r - \frac{5}{2} \). 7. Divide the leading term: \( 7r \div r = 7 \).