Y=(H)exp(Ax)+(K)exp(Bx)+F+Gx is the general solution of second order linear deq:(y'')+(-10y')+(21y)=(6) +(2)x

subject to zero initial conditions, i.e when x=0, y=0 and y'=0
find A,B,F,G,H,K

y = H e^(Ax) + K e^(Bx) + F + Gx

If you solve associated homogeneous equation: y'' − 10y' + 21y = 0
by finding solutions to characteristic equation, you get:
r² − 10r + 21 = 0
(r − 3) (r − 7) = 0
r = 3, r = 7

So we know that A = 3, B = 7

Solution to homogeneous equation: yh = H e^(3x) + K e^(7x)
Particular solution for non-homogeneous equation: yp = F + Gx

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yp = F + Gx
yp' = G
yp'' = 0

Now we plug these values of yp, yp', and yp'' into differential equation:

yp'' − 10yp' + 21yp = 6 + 2x
0 − 10G + 21 (F + Gx) = 6 + 2x
(21F−10G) + (21G)x = 6 + 2x

Matching coefficients, we get:
21G = 2 -----> G = 2/21
21F−10G = 6
21F−20/21 = 6 -----> F = 146/441

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y = H e^(3x) + K e^(7x) + 146/441 + 2/21 x
y' = 3H e^(3x) + 7K e^(7x) + 2/21

y(0) = 0 ------> H + K + 146/441 = 0
y'(0) = 0 -----> 3H + 7K + 2/21 = 0
H = −5/9
K = 11/49

y = −5/9 e^(3x) + 11/49 e^(7x) + 146/441 + 2/21 x

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