We first solve the homogeneous equation

\(2y" + 3y' − 2y = 0\)

Its characteristic equation is

\(2r^{2} + 3r - 2 = 0 \Rightarrow r = (=3 \pm \frac{\sqrt{9+16}}{4}=\frac{-3 \pm 5}{4}\)

Thus, \(r = -2\ or\ r = \frac{1}{2}\) the solution of the homogeneous equation is

\(y_{H} = C_{1}e^{-2x} + C_{2}e^{x/2}\)

Now we need to find some particular solution y,, a solution such that

\(2y"_{p}, + 3y'_{p} - 2y_{p} = 14x^{2} - 4x - 11\) (1)

We will try with polynomial

\(y_{p} = Ax^{2} + Bx + C\)

We have that

\(y'_{p} = 2Ax + B\)

\(y"_{p} = 2A\)

So, using (1),

\(2A + 2Ax + B + Ax^{2} + Bx + C = 14x^{2} - 4x - 11 \Rightarrow\)

\Rightarrow Ax^{2} + (B + 2A)x + (2A + B + C) = 14x^{2} - 4x - 11\)

So, we have a system of equations

\(A = 14\)

\(2A + B = -4\)

\(2A + B + C = -11\)

Thus,

\(A = 14, B = -32, C = -7,\)

so \(y_{p} = 14x^{2} - 32x - 7\)

Finally, the solution of the initial equation is given by

\(y = y_{H} + y_{p} = C_{1}e^{-2}x + C_{2}e^{x/2} + 28x - 32\)

Now use that

\(0 = y(0) = C_{1} + C_{2} - 7\)

and

\(0 = y'(0) = -2C_{1} + \frac{1}{2} C_{2} - 32\)

So, the system is

\(C_{1} + C_{2} = 7\)

\(-2C_{1} + \frac{1}{2} C_{2} = 32\)

Multiply the first equation by 2 and add it to the second equation:

\(\frac{5}{2}C_{2} = 46 \Rightarrow C_{2} = 18.4\)

Plug this into the first equation to get

\(C_{1} + 18.4 = 7 \Rightarrow C_{1} = -11.4\)

Finally, the solution is

\(y = -11.4e^{-2}x + 18.4e^{x/2} + 14x^{2} - 32x - 7\)

\(2y" + 3y' − 2y = 0\)

Its characteristic equation is

\(2r^{2} + 3r - 2 = 0 \Rightarrow r = (=3 \pm \frac{\sqrt{9+16}}{4}=\frac{-3 \pm 5}{4}\)

Thus, \(r = -2\ or\ r = \frac{1}{2}\) the solution of the homogeneous equation is

\(y_{H} = C_{1}e^{-2x} + C_{2}e^{x/2}\)

Now we need to find some particular solution y,, a solution such that

\(2y"_{p}, + 3y'_{p} - 2y_{p} = 14x^{2} - 4x - 11\) (1)

We will try with polynomial

\(y_{p} = Ax^{2} + Bx + C\)

We have that

\(y'_{p} = 2Ax + B\)

\(y"_{p} = 2A\)

So, using (1),

\(2A + 2Ax + B + Ax^{2} + Bx + C = 14x^{2} - 4x - 11 \Rightarrow\)

\Rightarrow Ax^{2} + (B + 2A)x + (2A + B + C) = 14x^{2} - 4x - 11\)

So, we have a system of equations

\(A = 14\)

\(2A + B = -4\)

\(2A + B + C = -11\)

Thus,

\(A = 14, B = -32, C = -7,\)

so \(y_{p} = 14x^{2} - 32x - 7\)

Finally, the solution of the initial equation is given by

\(y = y_{H} + y_{p} = C_{1}e^{-2}x + C_{2}e^{x/2} + 28x - 32\)

Now use that

\(0 = y(0) = C_{1} + C_{2} - 7\)

and

\(0 = y'(0) = -2C_{1} + \frac{1}{2} C_{2} - 32\)

So, the system is

\(C_{1} + C_{2} = 7\)

\(-2C_{1} + \frac{1}{2} C_{2} = 32\)

Multiply the first equation by 2 and add it to the second equation:

\(\frac{5}{2}C_{2} = 46 \Rightarrow C_{2} = 18.4\)

Plug this into the first equation to get

\(C_{1} + 18.4 = 7 \Rightarrow C_{1} = -11.4\)

Finally, the solution is

\(y = -11.4e^{-2}x + 18.4e^{x/2} + 14x^{2} - 32x - 7\)