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

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

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

### Relevant Questions

Using the existence and uniqueness theorem for second order linear ordinary differential equations, find the largest interval in which the solution to the initial value is certain to exist.
$$t(t^2-4)y''-ty'+3t^2y=0, y(1)=1 y'(1)=3$$
Give the correct answer and solve the given equation $$xy′′ − y′ = (3 + x)x^{2}e^{x}$$
Solve the second order linear differential equation using method of undetermined coefficients
$$3y''+2y'-y=x^2+1$$
Solve for the general solution of the given special second-ordered differential equation
$$xy''+x(y')^2-y'=0$$
Verify that the given functions form a basis of solutions of the given equation and solve the given initial value problem.
$$4x^2-3y=0,\ y(1)=3,\ y'(1)=2.5,$$ the basis of solution are $$y_1=x^{-\frac{1}{2}}$$ and $$y_2=x(\frac{3}{2})$$
Solve the differential equation:
$$y''+6y'+12y=0$$
A dynamic system is represented by a second order linear differential equation.
$$2\frac{d^2x}{dt^2}+5\frac{dx}{dt}-3x=0$$
The initial conditions are given as:
when $$t=0,\ x=4$$ and $$\frac{dx}{dt}=9$$
Solve the differential equation and obtain the output of the system x(t) as afunction of t.
$$x_1'=2x_1+x_2$$
$$x_2'=4x_1-x_2$$
$$y''-4y'+9y=0,\ \ y(0)=0,\ \ y'(0)=-8$$
Solve the linear equations by considering y as a function of x, that is, $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}.{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}}-{\left\lbrace{2}\right\rbrace}{\frac{{{x}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}{\left\lbrace{y}\right\rbrace}={\left\lbrace{x}\right\rbrace}^{{{2}}}$$