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Additional resources for Real Functions in One Variable Examples of Simple Differential Equations II Calculus Analyse 1c-5
Example text
K2 K Then a rearrangement gives 1 − e2 x2 + 2e 1 x + y2 = 2 , K K hence by division by 1 − e2 > 0 and addition of some convenient term, x2 + 2 · e 1 · 1 − e2 K e 1 ·x+ · 1 − e2 K e 1 1 1 · 2+ · 2 2 1−e K 1−e K which immediately is reduced to x+ 2 1 e · 2 1−e K + √ + 1 y2 1 − e2 2 = (12) 2 , y2 = 2 1 − e2 1 2 (1 − e2 ) K 2 , or (x + a)2 + y2 = c2 b2 where a= e 1 · , 2 1−e K 1 − e2 b= and c = 1 . (1 − e2 ) K This equation describes an ellipse with half axes c= 1 (1 − e2 ) K and bc = 1 √ . K 1 − e2 3) When e = ±1, it follows from (11) that x2 + y 2 = 1 2e · x + x2 , − K2 K which is reduced to the parabolic equation −2 1 e x + 2 = y2 , K K e = ±1.
12)] x− 1 e e2 − 1 K 2 − √ y2 e2 −1 2 = 1 (e2 2 − 1) K 2 , which is the equation of a hyperbola. 36 Find the complete solution of the differential equation d2 x dx +8 + 15 x = e−4t , dt2 dt t ∈ R. A. Linear inhomogeneous differential equation of second order and of constant coefficients. D. Find the roots of the characteristic equation and then find the solutions of the homogeneous equation. A particular solution can either be found by guessing or by the solution formula. I. The characteristic equation R2 + 8R + 15 = (R + 3)(R + 5) = 0 has the two simple roots R = −3 and R = −5.
1 This is equal to et for c = , hence a particular solution is given by 2 1 2 t x= t e. 2 3) a) If R = 1 is a root of the characteristic equation, then c · et is a solution of the corresponding homogeneous. Therefore, it never can be a solution of the inhomogeneous equation, if the right hand side is = 0. b) If the characteristic polynomial is R2 + a1 R + a0 = (R − 1)(R − a0 ), a0 = 1, then a1 = −1 − a0 . If we put x = c t et , then we get dx d2 x = c t e t + c et , = c t et + 2 c et . dt dt2 Then by insertion into the left hand side of the equation d2 x dx + a0 x + a1 2 dt dt d2 x dx = − (a0 + 1) + a0 x dt2 dt = c(t + 2)et − (a0 + 1)c(t + 1)et + a0 c t et = c {1 − (a0 + 1) + a0 } t et + c (2 − a0 − 1) et = c(1 − a0 )et .