Dtjytyk

I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.

The problem:

$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.

I understand how to find derivatives if I know the function, but this is stumping me.

Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation
$$
y^{primeprime} + y^prime - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
begin{align}label{eq:1}tag{1}
y^prime(x) = 2Ax + B quad text{and} quad y^{primeprime}(x) = 2A.
end{align}
So, if $y$ solves the differential equation, then
begin{align}
x^2 = y^{primeprime} + y^prime - 2y &= 2A + (2Ax+B) -2left(Ax^2 + Bx + C right)label{eq:2}tag{2}\
&= -2Ax^2 + (2A-2B)x + left( 2A + B -2Cright)label{eq:3}tag{3}.
end{align}
Comparing the coefficients of this equation, we find that
begin{align*}
begin{cases}
-2A = 1,\
2A-2B = 0,\
2A+B-2C = 0.
end{cases}
end{align*}
Solving this system will give the explicit values of $A,B$ and $C$ required.

If you plug the quadratic polynomial into your differential equation you will get

$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2.$$

Compare the coefficients and determine $A$, $B$, $C$.

With

$y = Ax^2 + Bx + C, tag 1$

we may substitute $y$, $y'$, and $y''$ into

$y'' + y' - 2y = x^2, tag 2$

viz,

$y' = 2Ax + B, tag 3$

$y'' = 2A, tag 4$

$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; tag 5$

we group together like powers of $x$:

$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, tag 5$

from which we infer

$-2A = 1, tag 6$

$2(A - B) = 0, tag 7$

$2A + B - 2C = 0; tag 8$

thus

$A = B = -dfrac{1}{2}, tag 9$

$C = -dfrac{3}{4}, tag{10}$

and of course

$y = -dfrac{1}{2}x^2 - dfrac{1}{2}x - dfrac{3}{4}. tag{11}$

We Check:

From (11),

$y' = -x - dfrac{1}{2}, tag{12}$

$y'' = -1, tag{13}$

$y'' + y' - 2y$
$= -1 - x - dfrac{1}{2} - 2(-dfrac{1}{2}x^2 - dfrac{1}{2}x - dfrac{3}{4}) = -1 - x - dfrac{1}{2} + x^2 + x + dfrac{3}{2} = x^2. tag{14}$

Consider $z^2+z-2=(z+2)(z-1)=-2(1+frac{1}{2}z)(1-z)$. Its inverse (as a formal power series) is
$$
-frac{1}{2}(1+z+z^2+dotsb)left(1-frac{1}{2}z+frac{1}{4}z^2+dotsbright)
=-dfrac{1}{2}-dfrac{1}{4}z-dfrac{3}{8}z^2+dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we can discard higher order terms):
$$
-dfrac{1}{2}x^2-dfrac{1}{4}2x-dfrac{3}{8}2=-dfrac{1}{2}x^2-dfrac{1}{2}x-dfrac{3}{4}
$$