the_Doctor111
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A particular solution to the differential equation y'' + 2y' + y = t^2 + 3?
is t^2 - 4t +7 in the answers, but i get t^2 - 4t + 9 so where am i going wrong...
y = Ay^2 + by + c
y' = 2Ay + B
y'' = 2A
2A + 2(2Ay + B) + Ay^2 +By + c = t^2 + 3
A(y^2) + (4A+B)y +(2A +2B+c) = t^2 + 0t + 3
A = 1
B + 4A = 0 → b = -4
C + 2B + 2A = 3 → b = 9
Sorry if it's just an arithmetical error I'm pretty tired
is t^2 - 4t +7 in the answers, but i get t^2 - 4t + 9 so where am i going wrong...
y = Ay^2 + by + c
y' = 2Ay + B
y'' = 2A
2A + 2(2Ay + B) + Ay^2 +By + c = t^2 + 3
A(y^2) + (4A+B)y +(2A +2B+c) = t^2 + 0t + 3
A = 1
B + 4A = 0 → b = -4
C + 2B + 2A = 3 → b = 9
Sorry if it's just an arithmetical error I'm pretty tired