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**Unformatted text preview: **18.03: Two practice final exams Operator Formulas • Exponential Response Formula: x p = Ae rt /p ( r ) solves p ( D ) x = Ae rt provided p ( r ) = 0. • Resonant Response Formula: If p ( r ) = then x p = Ate rt /p ( r ) solves p ( D ) x = Ae rt provided p ( r ) = 0. • Exponential Shift Law: p ( D )( e rt u ) = e rt p ( D + rI ) u . Properties of the Laplace transform ∞ 0. Definition: L [ f ( t )] = F ( s ) = f ( t ) e − st dt for Re s >> 0. − 1. Linearity: [ af ( t ) + bg ( t )] = aF ( s ) + bG ( s ). L 2. Inverse transform: F ( s ) essentially determines f ( t ). 3. s-shift rule: [ e at f ( t )] = F ( s − a ). L 4. t-shift rule: L [ f a ( t )] = e − as F ( s ), f a ( t ) = f ( t − a ) if t > a . if t < a 5. s-derivative rule: [ tf ( t )] = − F ( s ). L 6. t-derivative rule: [ f ( t )] = sF ( s ) − f (0+) L [ f ( t )] = s 2 F ( s ) − sf (0+) − f (0+) L if we ignore singularities in derivatives at t = 0. t L 7. Convolution rule: [ f ( t ) ∗ g ( t )] = F ( s ) G ( s ), f ( t ) ∗ g ( t ) = f ( τ ) g ( t − τ ) d τ . 1 8. Weight function: [ w ( t )] = W ( s ) = , w ( t ) the unit impulse response. L p ( s ) Formulas for the Laplace transform 1 1 n ! L [1] = L [ e at ] = s − a [ t n ] = s n +1 s L s ω L [cos( ωt )] = s 2 + ω 2 L [sin( ωt )] = e s 2 + ω 2 − as [ u a ( t )] = L [ δ a ( t )] = e − as L s where u ( t ) is the unit step function u ( t ) = 1 for t > 0, u ( t ) = for t < 0. Fourier series a f ( t ) = + a 1 cos( t ) + b 1 sin( t ) + a 2 cos(2 t ) + b 2 sin(2 t ) + ··· 2 π π 1 1 a m = f ( t ) cos( mt ) dt, b m = f ( t ) sin( mt ) dt π π π − π π − π cos( mt ) cos( nt ) dt = sin( mt ) sin( nt ) dt = for m = n − π π − π π 2 cos ( mt ) dt = sin 2 ( mt ) dt = π for m > − π − π If sq( t ) is the odd function of period 2 π which has value 1 between and π , then 4 sin(3 t ) sin(5 t ) sq( t ) = sin( t ) + + π 3 5 + ··· Variation of parameters The solution to u ˙ = A u + q ( t ) is given by u = Φ( t ) Φ( t ) − 1 q ( t ) dt where Φ( t ) is any fundamental matrix for A . (In fact this true even if the coeﬃcient matrix A = A ( t ) is nonconstant. The 1 × 1 case was studied early on.) Defective matrix formula If A is a defective 2 × 2 matrix with eigenvalue λ 1 and nonzero eigenvector v 1 , then you can solve for w in ( A − λ 1 I ) w = v 1 and u = e λ 1 t ( t v 1 + w ) is a solution to u ˙ = A u . Exam I 1. (a) A vial of pure Kryptonite undergoes radioactive decay, but Lex Luthor keeps it continually resupplied at the rate of q ( t ) grams per hour. Set up the ODE describing the number x ( t ) of grams of Kryptonite in the vial. Your answer will involve q ( t ) and an as yet undetermined decay rate....

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