Q.367. To find coefficients of a quadratic equation given a condition.

Question 367.
Using y = Ax^2 + Bx + C,  how would you solve for y'' - 4y' + 3y = 9x^2 ?

Answer 367.
y = Ax^2 + Bx + C
=> y' = 2Ax + B
=> y" = 2A
=> y" - 4y' + 3y = 9x^2
=> 2A - 4 (2Ax + B) + 3(Ax^2 + Bx + C) = 9x^2
=> 3Ax^2 + (3B - 8A)x + 2A - 4B + 3C = 9x^2
Comparing coefficients of x^2, 3A = 9 => A = 3,
Comparing coefficients of x, 3B - 8A = 0 => B = (8/3)A = 8 and
comparing the constant terms, 2A - 4B + 3C = 0 => C = (1/3) (4B - 2A) = 26/3
Answers:
A = 3,
B = 8 and
C = 26/3
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Verification:
y = 3x^2 + 8x + 26/3
=> y' = 6x + 8
=> y" = 6
=> y''- 4y' + 3y
= 6 - 4 (6x + 8) + 3 (3x^2 + 8x + 26/3)
= 6 - 24x - 32 + 9x^2 + 24x + 26
= 9x^2 (confirmed).

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