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FP2 Second Order Differentials

My question is about in what scenario to increase the power of the particular integral you choose. In the book (Edexcel FP2) it says if the particular integral is in the complementary function then you increase the power of xx or tt or whatever the variable is.

I thought this was okay until I came across this example:

d2ydt2+2dydt+2y=2e−t\frac{d^2y}{dt^2} + 2 \frac{dy}{dt} + 2y = 2e^{-t}

I need to find the general solution so did the standard "use the auxiliary equation to find the complementary function and then find the particular integral and put the two together to make the general solution".

So complementary function = e−t(Acos(t)+Bsin(t))e^{-t}(Acos(t) + Bsin(t))

I saw that there is the term e−te^{-t} in both the given differential equation and also in the complementary function so I assumed I had to increase the power of tt so I tried y=λte−ty = \lambda te^{-t} and differentiated twice to get d2ydt2\frac{d^2y}{dt^2} and dydt\frac{dy}{dt} and then substituted back into the differential to try and solve for λ\lambda but I just get λ=0\lambda = 0. Then I increased the power again and used y=λt2e−ty = \lambda t^{2}e^{-t} but this time the equations I got contradictory values of λ\lambda.

Then I resorted to looking at the mark scheme and saw that they used y=λe−ty= \lambda e^{-t} and it all worked when I used it.

Why does the rule not work? Or is it because the term in the complementary function is e−tcos(t)e^{-t}cos(t) and e−tsin(t)e^{-t}sin(t) and not actually y=λe−ty= \lambda e^{-t}?

p.s. apologies for the slightly waffley question, I got overexcited whilst trying LaTeX :colondollar:
Original post by Windows7Pro
Or is it because the term in the complementary function is e−tcos(t)e^{-t}cos(t) and e−tsin(t)e^{-t}sin(t) and not actually y=λe−ty= \lambda e^{-t}?



Essentially what you put: the particular integral has terms of e−tcos(t)e^{-t}cos(t) and e−tsin(t)e^{-t}sin(t) and not e−te^{-t} so no need to times it by t ...

If the solution was Ae−tAe^{-t} + Be−ktBe^{-kt} then you would need to multiply by t as it has one root of -1

If the solution was e−t(A+Bt)e^{-t}( A+Bt) then you'd multiply by t2t^{2} as it has repeated roots of -1

Hope this helps!
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Windows7Pro
OP
Ahhhhh I see I see. Thanks for confirming that.

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