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QUESTION

Find the complementary function ONLY of
$\ds\frac{d^2x}{dt^2}-5\frac{dx}{dt}+4x=e^t$.

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ANSWER


To find the complementary function we need to solve
$\ds\frac{d^2x}{dx^2}-5\frac{dx}{dt}+4x=0$\\ The auxiliary
equation is $m^2-5m+4=(m-4)(m-1)=0$ so $m=1,4$\\ The complementary
function is $x=Ae^t+Be^{4t}$



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