 # Solve the Differential equations 2y′′ + 3y′ − 2y = 14x^{2} − 4x − 11, y(0) = 0, y′(0) = 0 Dottie Parra 2020-11-02 Answered
Solve the Differential equations $2y\prime \prime +3y\prime -2y=14{x}^{2}-4x-11,y\left(0\right)=0,y\prime \left(0\right)=0$
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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
$$y = -11.4e^{-2}x + 18.4e^{x/2} + 14x^{2} - 32x - 7$$