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## Homework Statement

Find a particular solution to the differential equation

-6y''+5y'-1y

=

-1t^2+1t+1e^2t

## Homework Equations

y = y(c) + y(p)

## The Attempt at a Solution

First solve the homogeneous equation -6y" + 5y' - y = 0.

The characteristic equation is -6r^2 + 5r - 1 = 0 --> (1-2r)(1-3r) = 0 --> r = 1/2 or r = 1/3

So y(c) = Ae^(t/2) + Be^(t/3)

Now using methods of undetermined coefficients we assume y(p) = Ct^2 + Dt + E + Fe^(2t)

--> y'(p) = 2Ct + D + 2Fe^(2t)

--> y"(p) = 2C + 4Fe^(2t)

Substituting into the equation yields

LHS

= -6[2C + 4Fe^(2t)] + 5[2Ct + D + 2Fe^(2t)] - [Ct^2 + Dt + E + Fe^(2t)]

= -12C - 24Fe^(2t) + 10Ct + 5D + 10Fe^(2t) - Ct^2 - Dt - E - Fe^(2t)

= (-C)t^2 + (10C - D)t + (5D - E - 12C) + (-15F)e^(2t)

Comparing the coefficients on the LHS and RHS we have

-C = -1 --> C = 1

10C - D = 1 --> D = 9

5D - E - 12C = 0 --> E = 33

-15F = 1 -- F = -1/15

So y(p) = t^2 + 9t + 33 - (1/15)e^(2t)

Hence y = Ae^(t/2) + Be^(t/3) + t^2 + 9t + 33 - (1/15)e^(2t)