Answer: [tex]A = \frac{-1}{2}\\B = \frac{-1}{2}\\C = \frac{-3}{4}[/tex]
Step-by-step explanation:
First we derive y two times [tex]y = Ax^{2} + Bx + C\\y' = 2Ax + B\\y'' = 2A[/tex]
Second we replace in the differential equation y' and y'':
[tex]2A + 2Ax + B - 2(Ax^{2} + Bx + C) = x^{2}[/tex]
grouping terms
[tex](-2A)x^{2} + (2A-2B)x + (2A + B -2C) = x^{2}[/tex]
The terms on the left side must be equal to the terms of the right side, so
[tex]1. (-2A) x^{2} = x^{2} \\2. (2A - 2B)x = 0\\3. (2A + B - 2C) = 0[/tex]
From 1.
[tex]A = \frac{-1}{2}[/tex]
From 2.
[tex]2A = 2B\\B = A\\B = \frac{-1}{2}[/tex]
From 3.
[tex]2A + B - 2C = 0\\2A + A - 2C = 0\\3A - 2C = 0\\C = \frac{3A}{2}, A = -\frac{1}{2}\\C = -\frac{3}{4}[/tex]
