Solve the Differential equations 2y′′ + 3y′ − 2y = 14x^{2} − 4x − 11, y(0) = 0, y′(0) = 0

Question
Solve the Differential equations \(2y′′ + 3y′ − 2y = 14x^{2} − 4x − 11, y(0) = 0, y′(0) = 0\)

Answers (1)

2020-11-03
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
\(y = -11.4e^{-2}x + 18.4e^{x/2} + 14x^{2} - 32x - 7\)
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