I. feladatsor (1) T¨ oltse ki az al´ abbi t´ abl´ azatot: Komplex sz´ am Val´ os r´esz K´epzetes r´esz Konjug´alt Abszol´ √ut´ert´ek 4 + 2i 4 2 4 − 2i 20 −3 + 4i −3 4 −3 − 4i √5 −5i − 1 −1 −5 5i − 1 26 −6i 0 −6 6i 6 √ 3 − 5i 3 −5 3 + 5i √34 3 + 2i 3 2 3 − 2i 13 −7i 0 −7 7i 7 (2) Adottak az al´ abbi komplex sz´ amok: z1 = 2 + 3i, z2 = 1 − i, z3 = 2i. Hat´arozza meg az al´abbiakat: (a) √ 3 + 2i (b) √ 3 − 2i √ (c) 3 − 2i (d) 1 + 4i (e) 13 (f) 13 + 2 (g) 5 + i (h) √ −1 − 5i (i) 2 + 2i (j) 5 − i (k) 5 − i (l) 26 √ 1 5 1 5 3 (m) 26 (n) − + i (o) − − i (p) − i 2 2 13 13 2 1 5 1 5 1 (r) − − i (s) − − i (t) −5 − i (q) − i 2 2 2 2 √2 442 (u) 2 (3) Legyen z1 = 3(cos 60◦ + i sin 60◦ ) valamint z2 = 12(cos 120◦ + i sin 120◦ ). Hat´arozza meg az al´abbiakat: (a) 3 (b) 12 (c) 36(cos 180◦ + i sin 180◦ ) 1 (d) (cos 300◦ + i sin 3000◦ ) 4 (e) 33 (cos 180◦ + i sin 180◦ ) √ 60◦ + k · 360◦ 60◦ + k · 360◦ 3 (f) 3(cos + i sin ) k = 0, 1, 2 (20◦ , 140◦ , 260◦ ) 3 3 √ 120◦ + k · 360◦ 120◦ + k · 360◦ 4 (g) 12(cos + i sin ) k = 0, 1, 2, 3 (30◦ , 120◦ , 210◦ , 300◦ ) 4 4 ◦ ◦ ◦ ◦ √ 180 + k · 360 180 + k · 360 (h) 36(cos + i sin ) k = 0, 1 (90◦ , 270◦ ) 2 2 (4) T¨ oltse ki az al´ abbi t´ abl´ azatot: Algebrai alak Trigonometrikus alak √ 1 +√i 2(cos 45◦ + i sin 45◦ ) ◦ ◦ 1 − 3i √2(cos 300 ◦+ i sin 300 ◦) −1 + i 2(cos 135 + i sin 135 ) 2 2(cos 0◦ + i sin 0◦ ) i 1(cos 90◦ + i sin 90◦ ) (5) Oldja meg az al´ abbi egyenleteket a komplex sz´amok halmaz´an: 8 1 (a) z = 13 + 13 i (b) z1 = √ 1 + 3i, z2 = 1 −√3i (c) z1 = 3 + 2i,z2 = − 3 (d) z1 = 3, z2 = −3 (e) z1 = 4i, z2 = −4i √ 135◦ + k · 360◦ 135◦ + k · 360◦ 6 + i sin ) k = 0, 1, 2 (45◦ , 165◦ , 285◦ ) (f) z = 2(cos 3◦ 3 √ 71, 57 + k · 360◦ 71, 57◦ + k · 360◦ 12 (g) z = 10(cos + i sin ) k = 0, 1, 2, 3, 4, 5 6◦ 6 ◦ ◦ ◦ ◦ ◦ (11, 93 , 71, 93 , 131, 93 , 191, 93 , 251, 93 , 311, 93 ) 90◦ + k · 360◦ 90◦ + k · 360◦ (h) z1,2,3 = (cos + i sin ) k = 0, 1, 2 (30◦ , 150◦ , 270◦ ) 3◦ 3 √ 270 + k · 360◦ 270◦ + k · 360◦ 3 z4,5,6 = 3(cos + i sin ) k = 0, 1, 2 (90◦ , 210◦ , 330◦ ) 3 3 90◦ + k · 360◦ 90◦ + k · 360◦ (i) z1,2,3,4 = (cos + i sin ) k = 0, 1, 2, 4 (22, 5◦ , 112, 5◦ , 202, 5◦ , 292, 5◦ ) 4 4 ◦ ◦ ◦ √ 45 + k · 360 45 + k · 360◦ 8 z5,6,7,8 = 2(cos + i sin ) k = 0, 1, 2, 3 (11, 25◦ , 101, 25◦ , 191, 25◦ , 281, 25◦ ) 4 4 √ 90◦ + k · 360◦ 90◦ + k · 360◦ 3 (j) z1,2,3 = 2(cos + i sin ) k = 0, 1, 2 (30◦ , 150◦ , 270◦ ) 3 3 q √ 45◦ + k · 360◦ 45◦ + k · 360◦ 3 z4,5,6 = 2 2(cos + i sin ) k = 0, 1, 2, 3 (15◦ , 135◦ , 255◦ ) 3 3 (6) V´egezze el a kijel¨ olt m˝ uveleteket: 1
(a) 2 + 3i 300◦ + k · 360◦ 300◦ + k · 360◦ (b) 2(cos + i sin ) k = 0, 1, 2 (100◦ , 220◦ , 340◦ ) 3 3 √ 3 2 (c) (cos 75◦ + i sin 75◦ ) 2 √ (d) −1 + 3i 1 5 (e) + i 13 13 (7) Oldja meg az al´ abbi egyenleteket a komplex sz´amok halmaz´an: (a) z = 2 + 7i 23 11 (b) z = − − i 13 13 2 (c) z = − i 7 5 7 (d) z = − − i 3 3 √ √ 3 1 3 1 (e) z1 = 0, z2 = i, z3 = − i, z4 = − − i 2 2√ 2 2√ 3 3 1 1 (f) z1 = 0, z2 = 1, z3 = − + i, z4 = − − i 2 2 2 2 (g) z1 = i, z2 = −i (h) z1 = 1 − i, z2 = 2 − i (8) Hat´ arozza√meg az al´ abbi komplex sz´ amok trigonometrikus alakj´at: ◦ ◦ (a) z = 2(cos 315 + i sin 315 ) ◦ (b) z = 2(cos i sin 315◦ ) √ 315 + ◦ (c) z = 2 2(cos 45 + i sin 45◦ ) (d) z = 4(cos 30◦ + i sin 30◦ ) (e) z = cos 60◦ + i sin 60◦
II. feladatsor (1) T¨ oltse ki az al´ abbi t´ abl´ atatot: f (x) g(x) f (g(x)) g(f (x)) f (f (x)) 3 sin(x) x3 sin(x3 ) sin (x) sin(sin(x)) √ √ √ √ sin(x) + x + 1 x sin( x) + x + 1 sin x + x + 1 sin(sin(x) + x + 1) + sin x + x + 1 + 1 1 1 1 1 3x + 1 3 x+2 + 1 1 x+2 3x + 1 + 2 x+2 + 2 4 4 4 cos(x) x cos(x ) cos (x) cos(cos(x)) sin(x) 2x sin(2x ) 2sin(x) sin(sin(x)) 3 lg(x) x3 + 2x − 1 lg(x3 + 2x − 1) lg (x) + 2 lg(x) − 1 lg(lg(x)) p √ √ arcsin(x) x arcsin( x) arcsin(x) arcsin(arcsin(x)) x ln(x) e x x ln(ln(x)) ln(x) x2 ln(x2 ) ln2 (x) ln(ln(x)) ln(x) ln(x) ln(ln(x)) ln(ln(x)) ln(ln(x)) (2) Hat´ arozza meg a val´ os sz´ amok legb˝ ovebb r´eszhalmaz´at, ahol az al´abbi f¨ uggv´enyek ´ertelmesek: (a) Df =] − ∞; −4[∪] − 4; −1] ∪ [5; ∞[ 3 (b) Df =] − 2; 2] \ {− } 2 (c) Df = R \ {3} (d) Df = R \ {−2} (e) Df =] − ∞; −11[∪]2; ∞[ (f) Df = [−1; ∞[ 1 (g) Df =] ; ∞[ 3 (h) H´ arom kik¨ ot´est kell tenni: • arccos(x) ´ertelmez´esi tartom´ anya: [−1, 1]. Jel¨olje ezt H1 . • ln(x) ´ertelmez´esi tartom´ anya: ]0, ∞[. Jel¨olje ezt H2 . • a nevez˝ o nem lehet nulla. Jel¨ olje ezt H3 . Ezekket: Df = H1 ∩ H2 \ H3 Ezeket k¨ ul¨ on-k¨ ul¨ on megoldjuk: • −1 ≤ x + 2 ≤ 1
\−2
−3 ≤ x ≤ −1 azaz H1 = [−3; −1] • x+4>0
\−4
x > −4 azaz H2 =] − 4; ∞[ • ln(x + 4) 6= 0 = ln 1 a logaritmus f¨ uggv´eny szigor´ u monotonit´asa miatt x + 4 6= 1
\−4
x 6= −3 azaz H3 = {−3} Vagyis: Df = H1 ∩ H2 \ H3 =] − 3; −1] (3) (4) Hat´ arozza meg az al´ abbi f¨ uggv´enyek inverz f¨ uggv´eny´et, ha l´etezik: x+6 (a) Df = R, Rf = R, Df = R, Rf = R, f (x) = 4
x2 − 6 2 (c) Df = R \ {−5}, Rf = R \ {−6}, Df = R \ {−6}, Rf = R \ {−5}, f (x) =
(b) Df = [−3; ∞[, Rf = [0; ∞[, Df = [0, ∞[, Rf = [−3, ∞[, f (x) =
3 −5 x+6
(d) Df = R, Rf =]3, ∞[, Df =]3, ∞[, Rf = R, f (x) = log2 (x − 3) + 7 5x−6 + 10 (e) Df =]5, ∞[, Rf = R, Df = R, Rf =]5, ∞[, f (x) = 2 (f) Df = R, Rf = [3, ∞[ de ´ıgy nem invert´alhat´o. Df = [4, ∞[, Rf = [3, ∞[, Df = [3, ∞[, Rf = [4, ∞[, √ f (x) = x − 3 + 4
2 (g) Df = R \ {4}, Rf = R \ {3}, Df = R \ {3}, Rf = R \ {4}, f (x) = +4 x − 3 1 x−π 1 2 sin +1 (h) Df = [− ; 1] = Rf , Rf = [0; 2π] = Df , f (x) = 3 3 2 π π (i) Df = R = Rf , Rf = [− ; 0] = Df , f (x) = 4tg 2x + +1 2 2 (j) Df = R, Rf = [−4; −2] de ´ıgy nem invert´alhat´o. Lesz˝ uk´ıt´es pl.: Df = [π; 5π] = Rf , Rf = [−4; −2] = Df , f (x) = 4 arccos(x + 3) + π (5) P´ aros, p´ araltlan vagy egyik sem az al´ abbi f¨ uggv´eny: (a) f (x) = e−x x2 , f (−x) = ex x2 azaz egyik sem. 2 2 (b) f (x) = e−x sin x, f (−x) = −e−x sin x, azaz p´aratlan. (c) f (x) = x2 + cos x + 2, f (−x) = x2 + cos x + 2, azaz p´aros.
III. feladatsor
5n − 1 4 9 5n + 4 5n − 16 sorozat k¨ ovetkez˝ o elemeit: a1 = , a2 = , an+1 = , an−3 = . 2n + 1 3 5 2n + 3 2n − 5 (2) Vizsg´ alja meg az al´ abbi sorozatokat monotonit´as ´es korl´atoss´ag szempontj´ab´ol: 1 (a) Szigor´ uan monoton cs¨ okken˝ o, K = a1 = k = lim an = 0. n→∞ 5 4 (b) Szigor´ uan monoton n˝ o, k = a1 = K = lim an = 1 n→∞ 3 1 1 (c) Se nem cs¨ okken, se nem n˝ o. K = k = − , lim an = 0 4 2 n→∞ 4 (d) A harmadik tagt´ ol kezdve szigor´ uan monoton cs¨okken˝o. K = a3 = 7 k = a2 = −10, lim an = n→∞ 3 (3) Adottak az al´ abbi konvergens sorozatok. Adja meg a hat´ar´ert´eket, ´es azt a k¨ usz¨obindexet, amelyn´el nagyobb index˝ u 1 tagjai a sorozatnak ε = -n´ al kisebb hib´ aval k¨ozel´ıtik a hat´ar´ert´eket! 100 (a) lim an = 4, n0 = 100 (b) lim an = −3, n0 = 1296 n→∞ n→∞ 4 3 (c) lim an = , n0 = 31 (d) lim an = , n0 = 380 n→∞ n→∞ 4 3 (4) Hat´ arozza meg az al´ abbi nevezetes sorozatok hat´ar´ert´eket: (a) lim an = 3 (b) lim an = 0 (c) lim an = ∞
(1) Adja meg az an =
n→∞
n→∞
n→∞
(d) lim an = ∞
(e) lim an = 1
(f) lim an = 0
(g) lim an = e−3
(h) lim an = ∞
(i) lim an = 0
(j) lim an = 1
(k) lim an = 0
(l) lim an = 1
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
(5) A nevezetes sorozathat´ ar´ert´ekek ismeret´evel ´es a m˝ uveletekre vonatkoz´o t´etelek seg´ıts´eg´evel hat´arozza meg mely sorozatok konvergensek ´es mi a hat´ ar´ert´ek¨ uk! (a) lim an = ∞ (b) lim an = ∞ n→∞
(c) lim an = ∞ n→∞
n→∞
(d) lim an = −∞ n→∞
(6) Hat´ arozza meg a k¨ ovetkez˝ o sorozatok hat´ ar´ert´ek´et: 5 3 (a) lim an = − (b) lim an = n→∞ n→∞ 7 5 7 (d) lim an = 0 (c) lim an = − n→∞ n→∞ 9 (e) lim an = 0 (f) lim an = 0 n→∞
(g) lim an = ∞ n→∞
(i) lim an = 2 n→∞
n→∞
(h) lim an = −∞ n→∞ 7 (j) lim an = n→∞ 3 (l) lim an = −1
1 n→∞ n→∞ 7 (7) Hat´ arozza meg a k¨ ovetkez˝ o sorozatok hat´ ar´ert´ek´et: 4 1 (a) lim an = √ (b) lim an = − 3 n→∞ n→∞ 3 5 15 (c) lim an = (d) lim an = 0 n→∞ n→∞ 16 (e) lim an = ∞ (f) lim an = 3 n→∞ n→∞ 1 2 (l) lim an = (k) lim an = √ n→∞ n→∞ 3 5 (8) Hat´ arozza meg a k¨ ovetkez˝ o sorozatok hat´ ar´ert´ek´et: 1 1 (a) lim an = − (b) lim an = n→∞ n→∞ 3 4 (c) lim an = ∞ (d) lim an = 0 (k) lim an = −
n→∞
n→∞
(e) lim an = −∞ (f) lim an = 0 n→∞ n→∞ 1 (g) lim an = (h) lim an = ∞ n→∞ n→∞ 32 (9) Hat´ arozza meg a k¨ ovetkez˝ o sorozatok hat´ ar´ert´ek´et:
(a) lim an = e2
(b) lim an = e2
(c) lim an = ∞
(d) lim an = 1
(e) lim an = e3
(f) lim an = e−4
(g) lim an = e3
(h) lim an = e−1
(i) lim an = e34
(j) lim an = e− 7
(k) lim an = 0
(l) lim an = 0
(m) lim an = ∞
(n) lim an = ∞
(o) lim an = 1
(p) lim an = e− 2
(q) lim an = 0
(r) lim an = e 3
(s) lim an = ∞
(t) lim an = 0
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞ n→∞
n→∞
n→∞
3
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
3
14
IV. feladatsor (1) Hat´ arozza meg az al´ abbi f¨ uggv´enyhat´ ar´ert´ekeket a f¨ uggv´eny grafikonj´anak ismeret´eben: (a) lim x3 = +∞, lim x3 = −∞ x→+∞
x→−∞
1 1 1 = 0, lim = 0, lim nem l´etezik (b) lim x→−∞ x x→+∞ x x→0 x 1 1 1 = 0, lim 2 = +∞ (c) lim 2 = 0, lim x→−∞ x2 x→+∞ x x→0 x √ √ (d) lim x nem l´etezik, lim+ x = 0 x→0
x→0
lim 2x = ∞, lim 2x = 0 x→+∞ x x→−∞ x 1 1 (f) lim = 0, lim = +∞ x→−∞ 3 x→+∞ 3 (g) lim lg x nem l´etezik, lim+ lg x = −∞, lim lg x = +∞ (e)
x→0
x→+∞
x→0
(h) lim log 21 x nem l´etezik, lim+ log 12 x = +∞, lim log 12 x = −∞ x→0
(i) (j)
x→+∞
x→0
lim sin x nem l´etezik, lim sin x nem l´etezik
x→+∞
x→−∞
lim tg x = +∞ π−
x→ 2
π π lim arcsin x = − , lim arcsin x nem l´etezik, lim arcsin x = + − 2 x→1 2 x→−1 x→1 π (l) lim arctg x = π2 , lim arctg x = x→+∞ x→−∞ 2 (m) lim + arccos x = π, lim arccos x nem l´etezik (k)
x→1
x→−1
(n)
lim arcctg x = 0, lim arcctg x = π
x→+∞
x→−∞
(o) ( x2 + 4 f (x) = 4x
ha x ≥ 1 ha x < 1
lim f (x) = ∞, lim f (x) = 0, lim f (x) nem l´etezik, lim f (x) = 8
x→+∞
x→−∞
x→1
x→2
(p) ( f (x) =
1
ha x < 2 x ha x ≥ 2
2
x √
lim f (x) = +∞, lim f (x) = +∞, lim f (x) nem l´etezik, lim f (x) = 3
x→+∞
x→0
x→2
x→9
(q) (
1 x 1 3 3 − x1
ha x ≥ −1 ha x < −1 11 1 1 lim f (x) = 0, lim f (x) = 1, lim f (x) = , lim f (x) = x→−∞ x→−1 x→0 3 x→10 3 (2) Hat´ arozza meg az al´ abbi f¨ uggv´enyek x0 -beli hat´ar´ert´ek´et: (a) lim f (x) = ∞, lim f (x) = −∞ f (x) =
x→−∞
(b)
x→∞
lim f (x) = −∞, lim f (x) = ∞
x→−∞
x→∞
3 3 (c) lim f (x) = − , lim f (x) = − x→−∞ 4 x→∞ 4 7 7 (d) lim f (x) = − , lim f (x) = − x→−∞ 2 x→∞ 2 (e) lim f (x) = 0, lim f (x) = 0 x→−∞
(f)
x→∞
lim f (x) = −∞, lim f (x) = ∞
x→−∞
x→∞
5 , lim f (x) = 2, lim f (x) nem l´etezik x→1 4 x→0 2 3 (h) lim f (x) = 0, lim f (x) = 0, lim f (x) nem l´etezik, lim f (x) = , lim f (x) = , lim f (x) nem l´etezik x→−∞ x→∞ x→0 x→1 x→2 3 4 x→4 1 (i) lim f (x) = , lim f (x) = 0, lim f (x) = −1, lim f (x) nem l´etezik x→0 x→2 x→3 3 x→1 (j) lim f (x) = nem l´etezik, lim f (x) = 0 (g)
lim f (x) = 1, lim f (x) = 1, lim f (x) =
x→−∞
x→∞
x→−1
x→−3
x→2
(k) lim f (x) = 0, lim f (x) = 0 x→−3
(l)
x→0
lim f (x) = −∞, lim f (x) = ∞
x→−∞
x→∞
1 (m) lim f (x) = , lim f (x) = 0 x→0 2 x→∞
(n)
lim f (x) = −1, lim f (x) = 1
x→−∞
x→∞
(3) Hat´ arozza meg az al´ abbi hat´ ar´ert´ekeket: √ √ 2x x √ √ = − 5 (b) lim √ =− 3 (a) lim √ x→0 x→0 5−x− 5+x 3 − 2x − 3 + 2x sin(2x) 2 sin 5x 5 (c) lim = (d) lim = x→0 x→0 7x 3x 3 7 5 tg 5x 5 sin 5x (e) lim = (f) lim = x→0 sin 7x x→0 4x 7 4 5x 4x e −1 5 e −1 4 (g) lim = (h) lim = x→0 x→0 sin(3x) 7x 7 2x 3 1 e −1 π (j) lim arctg (i) lim e x = 1 = x→∞ x→0 2x 4 5x−1 2x+1 5 1 x + 4 (k) lim 1 − = e− 2 (l) lim = e−4 x→∞ x→∞ x + 6 2x + 3 x+4 3x−2 1 27 3x + 1 5x − 2 = e− 3 = e− 5 (m) lim (n) lim x→∞ 3x + 2 x→∞ 5x + 7 x+1 3x 2x + 5 x+1 =0 (p) lim =0 (o) lim x→∞ 7x − 3 x→∞ 2x + 3 2x+3 4x−1 3x + 2 5x + 2 (q) lim =∞ (r) lim =∞ x→∞ 2x − 1 x→∞ x+3 2 2x+3 x2 +3 2x + 2 4x − 3 (s) lim = 1 (t) lim =∞ x→∞ 2x2 − 1 x→∞ 2x − 1 2 2 2x +3 2 2x+3 4 3x + 2 3x + 2x + 20 2 = e (v) lim = e3 (u) lim x→∞ x→∞ 3x2 − 1 3x2 + 20 1 1 (w) lim e x = ∞ (x) lim e x = 0 x→0+ x→0− 1 π 1 π (y) lim arctg = (z) lim arcsin 1 − = x→∞ x→1+ x−1 2 x 2
V. feladatsor (1) Adja meg az x0 helyhez tartoz´ o differenciah´ anyados f¨ uggv´enyt, a differenci´alh´anyados ´ert´ek´et a defin´ıci´ o alapj´ an: 5−5 0 (a) f (x) = 5, x0 = 2, x0 = −3 d2 (x) = = 0 f (2) = 0 x−2 5−5 = 0 f 0 (−3) = 0 d−3 (x) = x+3 1 −1 1 0 1 1 f (2) = − (b) f (x) = , x0 = 2, x0 = −3 d2 (x) = x 2 = − x x−2 2x 4 1 1 1 0 1 x + 3 d−3 (x) = = f (−3) = − 3x 4 √ x+3 (c) f (x) = x + x, x0 = −3, x0 = 2, x0 = 4 x0 = −3 √ -ban a f¨ uggv´ √ eny nincs ´ertelmezve. √ √ x+x− 2−2 x− 2 1 1 √ f 0 (2) = 1 + √ d2 (x) = =1+ =1+ √ x−2 x−2 x+ 2 2 2 √ √ √ √ x+x− 4−4 x− 4 1 1 √ f 0 (4) = 1 + d4 (x) = =1+ =1+ √ x−4 x−4 4 x+ 4 (2) Adja meg az al´ abbi f¨ uggv´enyek szel˝ o egyeneseinek egyenlet´et, valamint az ´erint˝o egyenesek egyenlet´et: (a) f (x) = 4x2 + 1, P1 (0, 1), P2 (−1, 5) e1 : y = 1, e2 : y = −8x − 3, sz : y = −4x + 1 1 1 1 1 2 1 1 1 (b) f (x) = , P1 (2, ), P2 (−3, − ) e1 : y = − x + 1, e2 : y = − x − , sz : y = x + x 2 3 4 9 3 6 6 (3) Hat´ arozza meg az al´ abbi f¨ uggv´enyek deriv´ altf¨ uggv´enyeit: 5 2 1 −3 1 0 (a) f (x) = x 3 − x 2 − 2 3 x 2 x 2 2 0 (b) f (x) = ln − 4 cos x 3 3 1 0 2 (c) f (x) = 6x + x ln 3 5 (d) f 0 (x) = 12x3 + sin x − 2 x 1 (e) f 0 (x) = cos x · tg x + sin · 2 cos x 1 (f) f 0 (x) = ex · ln x + ex · x 1 1 (g) f 0 (x) = · arcsin x + lg x · √ x ln 10 1 − x2 √ 1 1 (h) f 0 (x) = √ lg x + x · x ln 10 2 x 1 0 4 (i) f (x) = 20x · arctg x + 4x5 · 1 + x2 1 0 (j) f (x) = · sin x + ln x · cos x x 1 0 (k) f (x) = cos2 x 1 (l) f 0 (x) = − 2 sin x ex sin x − ex cos x (m) f 0 (x) = sin2 x 1 ln x − tg x x1 2 (n) f 0 (x) = cos x 2 ln x (2x ln 2 − 4x3 )(x + 2) − (2x − x4 ) 0 (o) f (x) = (x + 2)2 √ 1 − 23 3 sin x − ( 3 x + 15)3 cos x 0 3x (p) f (x) = 9 sin2 x 1 x − + 4) − cgtx · ex (e 2 (q) f 0 (x) = sin x (ex + 4)2 0 3 (r) f (x) = cos(x ) · 3x2 (s) f 0 (x) = 3(sin x)2 cos x 1 (t) f 0 (x) = 2 · 2x x +4 2 1 1 (u) f 0 (x) = ex − x +3 · (2x + 2 ) x 2 ln(x2 + cos x) 2x − sin x (v) f 0 (x) = etg x·ln(x +cos x) · + tg x cos2 x2 + cos x 0 x (w) f (x) = x · (ln x + 1)
cos x (x) f 0 (x) = (sin x)cos x · ( cos x − ln(sin x) sin x) sin x ! √ √ 1 − 12 (2x + x2 ) − x + 3(2x ln 2 + 2x) x+3 0 2 (x + 3) · (y) f (x) = cos 2x + x2 (2x + x2 )2 ln(x3 + 3) 3x2 (z) f 0 (x) = (x3 + 3)arctg x arctg x + x3 + 3 1 + x2 1 (aa) r0 (ϕ) = 4ϕ ln 4 · ln ϕ + 4ϕ ϕ (ab) V 0 (t) = 6t5 · cos(3t) − t6 sin(3t) · 3 (ac) h0 (t) = p5 + 1, p val´ os param´eter (ad) h0 (p) = t5 · p4 − 7p6 , t val´ os param´eter ! 1 √ + 3x ln 3 √ √ √ 1 2 x x x+4 x x+4 x 0 √ + ctg x · ( x + 3 ) (x + 4) + ln( x + 3 ) (ae) f (x) = − 2 · ( x + 3 ) x + 3x sin x (4) Hat´ arozza meg az al´ abbi hat´ ar´ert´ekeket: arcsin(5x) 5 x2 + 2 1 lim = = lim lim (x2 · ln(x)) = 0 lim xx = 1 2 x→∞ x→0 3x x→0+ x→0+ 3x x 3 + x + 11 3 1 1 1 1 x − sin x 1 1 lim 1+ = e lim − =− lim =0 lim − =0 x→−∞ x x − 1 ln x 2 x→0 x sin x x sin x x→1+ x→0+ x x−1 ln 2 1 1 arctan x 1 − cos x = lim = lim =1 lim e x−x2 = e−1 lim 2 x→2 x − 2 x→0 x→1 x→0 x 2 2 x √ 3 2x 3 =∞ lim 2x · ctg 3x = lim lim (sin x)x = 1 lim (ctg x) x = 1 x→∞ x2 x→0 x→0+ 2 x→0+ 1 5x lim (x − ln(x2 + 1)) = ∞ lim ln x · tg x = 0 lim =∞ lim (ex + x) x = e2 x→∞ x→∞ ln x x→0+ x→0+ (5) ´Irja fel az al´ abbi f (x) f¨ uggv´enyek m meredeks´eg˝ u ´erint˝oinek egyenlet´et: (a) x1 = 1, e1 : y = 6x + 25, x2 = 2, e1 : y = 6x + 24 5 36 7 72 (b) x1 = − , e1 : y = 10x + , x2 = − , e1 : y = 10x + 3 3 3 3 1 1 (c) x1 = 1, e1 : y = x + 0, 6482, x2 = −4, e1 : y = x − 0, 5329 26 26 (6) Hol n¨ ovekv˝ o, hol cs¨ okken˝ o, hol van lok´ alis sz´els˝o´ert´eke ´es milyen a sz´els˝o´ert´ek jellege az f (x) f¨ uggv´enynek, ha a deriv´ alt f¨ uggv´enye az al´ abbi: (a) f 0 (x) = (x + 4)(x − 3)2 ] − ∞; −4[ −4 ] − 4; 3[ 3 ]3, ∞[ f 0 (x) 0 + 0 + f (x) & lok. min. % % (x + 5)2 (x − 1) (b) f 0 (x) = x−3 ] − ∞; −5[ −5 ] − 5; 1[ 1 ]1; 3[ 3 ]3, ∞[ f 0 (x) + 0 + 0 X + f (x) % % lok. max. & % (x − 2)2 (x − 5) (c) f 0 (x) = ln x · (x + 3)(x − 4) ]0; 1[ 1 ]1; 2[ 2 ]2; 4[ 4 ]4,5[ 5 ]3, ∞[ f 0 (x) 0 + 0 + X 0 + f (x) & lok. min. % % & lok. min. % (7) Hol konvex, hol konk´ av, hol van inflexi´ os pontja az f (x) f¨ uggv´enynek, ha a m´asodik deriv´alt f¨ uggv´enye az al´ abbi: (a) f 00 (x) = (x + 4)(x − 3)2 ] − ∞; −4[ −4 ] − 4; 3[ 3 ]3, ∞[ 00 f (x) 0 + 0 + f (x) _ infl. pont ^ ^ (x + 4)3 (x − 2)2 (b) f 00 (x) = x+1 ] − ∞; −4[ −4 ] − 4; −1[ -1 ] − 1; 2[ 2 ]2, ∞[ f 00 (x) + 0 X + 0 + f (x) ^ infl. pont _ ^ ^ √ x(x − 4) 00 (c) f (x) = · 2x (x − 1)2 (x − 2) 0 ]0; 1[ 1 ]1; 2[ 2 ]2; 4[ 4 ]4, ∞[ f 00 (x) 0 + X + X 0 + f (x) ^ ^ _ infl. pont ^
(8) V´egezzen teljes f¨ uggv´enyvizsg´ alatot az al´ abbi f¨ uggv´enyeken: (a) 1. Df = R √ √ 2. z´erushelye van a − 3, 0, 3 pontokban 3. p´ aratlan, f (0) = 0 4. lim f (x) = ∞, lim f (x) = −∞. x→−∞
x→∞
5. f 0 (x) = 3 − 3x2 , a deriv´ alt z´erushelyei (lehets´eges lok´alis sz´els˝o´ert´ekhelyek): −1, 1. 6. x f 0 (x) f (x)
] − ∞, −1[ − &
−1 0 lok. min. f (−1) = −2
] − 1, 1[ + %
1 0 lok. max f (1) = 2
]1, ∞[ − &
7. f 00 (x) = −6x, a m´ asodik deriv´ alt z´erushelyei (lehets´eges inflexi´os pontok): 0. 8. x f 00 (x) f (x)
] − ∞, 0[ + ^ konvex
0 0 inflexi´os pont f (0) = 0
]0, ∞[ − _ konk´av
10. Rf = R. (b) 1. Dg = R. 2. nincs z´erushelye. 3. p´ aros, f (0) = 1 4. lim g(x) = lim g(x) = 0. x→−∞ 0
5. g (x) =
x→∞
−2x
2,
(1 + x2 )
a deriv´ alt z´erushelye az x = 0.
6. x f (x) f (x) 0
] − ∞, 0[ + %
]0, ∞[ − &
0 0 lok. max. f (0) = 1
√ √ 2(3x2 − 1) , a m´ asodik deriv´alt z´erushelyei az x = −1/ 3, 1/ 3 pontok. 2 3 (1 + x ) √ √ √ √ √ √ x ] − ∞, −1/ 3[ −1/ 3 ] − 1/ 3, 1/ 3[ 1/ 3 ]1/ 3, ∞[ f 00 (x) + 0 − 0 + f (x) ^ inflexi´ o s pont _ inflexi´ o s pont ^ 1 3 1 3 konvex f −√ = konk´av f −√ = konvex 4 4 3 3
7. g 00 (x) =
10. Rg =]0; 1] (c) h(x) = 4x2 + x1 1. A h f¨ uggv´eny mindenhol ´ertelmezett, kiv´eve az x = 0 pontot. Dh = R \ {0}. 1 2. Z´erushely: x = − √ 3 . 4 3. h(−x) = 4x2 − x1 , nem p´ aros, nem p´ aratlan. h(0) nincs ´ertelmezve. 4. lim h(x) = ∞, lim h(x) = −∞, lim h(x) = ∞, lim h(x) = ∞ x→−vv
5. h0 (x) = 8x − 6.
x→0−
1 x2 .
x h (x) h(x) 0
7. h00 (x) = 8 + 8. 10. Rh = R
2 x3 ,
x→∞
x→0+
h0 (x) = 0 akkor, ha x = 12 . ] − ∞, 0[ − &
1 h00 (x) = 0, ha x = − √ 3 4
0 X
]0, 12 [ − &
1 2
0 lok. min. h( 12 ) = 3
] 21 , ∞[ + %
1 ] − ∞, − √ 3 [ 4 + ^
x 00 h (x) h(x)
1 −√ 3 4 0 inflexi´os pont 1 h(− √ 3 ) = 0 4
]−
1 √ 3 , 0[ 4
− _
0 X
]0, ∞[ + ^
2
(d) i(x) = x2x+4 1. Di = R. 2. Z´erushely x = 0. 2 3. i(−x) = x2x+4 = i(x) azaz a f¨ uggv´eny p´aros, i(0) = 0 4. lim i(x) = 1, lim i(x) = 1. x→−∞
5. i0 (x) = 6.
x→∞ 8x 0 (x4 +4)2 , i (x)
= 0, ha x = 0 x i0 (x) i(x)
7. i00 (x) = 8.
8(4−3x2 ) 00 (x2 +4)3 , i (x)
0 0 lok. min. i(0) = 0
√
= 0 ha x1 = − 2 3 3 , x2 = √
x i00 (x) i(x)
] − ∞, 0[ − &
] − ∞, − 2 3 3 [ − _
√ 2 3 3
√
−233 0 inflexi´ o s pont √ i(− 2 3 3 ) = 14
]0, ∞[ + %
]−
√ √ 2 3 2 3 , 3 3 [
+ ^
√ 2 3 3
0 inflexi´ o s pont √ i( 2 3 3 ) = 14
√
] 2 3 3 , ∞[ − _
10. Ri = [0; 1[ (e) j(x) = x2 e−x 1. Dj = R 2. Z´erushely: x = 0 3. j(−x) = x2 ex nem p´ aros, nem p´ aratlan. j(0) = 0. 4. lim j(x) = ∞, lim j(x) = 0 x→−∞
x→∞
5. j 0 (x) = e−x (2x − x2 ), j 0 (x) = 0 ha x1 = 0, x2 = 2 6. x j 0 (x) j(x)
] − ∞, 0[ − &
0 0 lok. min. j(0) = 0
]0, 2[ + %
2 0 lok. max. j(2) = e42
]2, ∞[ − &
√ √ 7. j 00 (x) = e−x (x2 − 4x + 2), j 00 (x) = 0, ha x1 = 2 − 2, x2 = 2 + 2. 8. √ √ √ √ √ x ] − ∞, 2 − 2[ 2− 2 ]2 − 2, 2 + 2[ 2+ 2 j 00 (x) + 0 − 0 j(x) ^ inflexi´ o s pont _ inflexi´ o √ √ s pont j(2 − 2) ≈ 0, 19 j(2 + 2) ≈ 0, 38
]2 +
√
2, ∞[ + ^
10. Rj =]0, ∞[ (f) k(x) = x lg(x) 1. Dk =]0; ∞[ 2. Z´erushely: x = 1 3. nem p´ aros, nem p´ aratlan, ui. negat´ıv x-re nincs ´ertelmezve. k(0) szint´en nincs ´ertelmezve. 4. lim k(x) = 0, lim k(x) = ∞ x→0+
x→∞
5. k 0 (x) = lg x + ln110 , k 0 (x) = 0 ha x = 1e 6. 7. k 00 (x) = x ln1 10 , k 00 (x)-nek nincs z´erushelye 8. 10. Rk =] − 0, 16, ∞[
x k (x) k(x) 0
]0, 1e [ − &
1 e
0 lok. min. k( 1e ) = −0, 16
] 1e , ∞[ + %
]0, ∞[ + ^
x k 00 (x) k(x) (g) l(x) = ln(x2 + 4) 1. Dl = R 2. Z´erushely nincs. 3. l(−x) = ln(x2 + 4) = l(x) p´ aros, l(0) = ln 4. 4. lim l(x) = ∞, lim l(x) = ∞ x→−∞
5. l0 (x) = 6.
x→∞ 2x 0 x2 +4 , l (x) = 0
ha x = 0 x l0 (x) l(x)
7. l00 (x) = 8.
8−2x2 00 (x2 +4)2 , l (x)
x l00 (x) l(x)
] − ∞, 0[ − &
0 0 lok. min. l(0) = ln 4
]0, ∞[ + %
= 0, ha x1 = −2, x2 = 2.
] − ∞, −2 − _
−2 0 inflexi´os pont l(−2) = ln 8
] − 2; 2 + ^
2 0 inflexi´os pont l(−2) = ln 8
]2; ∞[ − _
10. Rl = [ln 4, ∞[ (h) m(x) = ln(x2 − 1) 1. Dm =] − ∞; −1[∪]1; √ ∞[ √ 2. Z´erushely: x1 = − 2, x2 = 2. 3. m(−x) = ln(x2 − 1) = m(x) p´ aros, m(0) nincs ´ertelmezve 4. lim m(x) = ∞, lim m(x) = ∞ lim m(x) = −∞, lim m(x) = −∞ x→−∞ 0
5. m (x) = 6.
x→∞
2x x2 −1 ,
x→−1−
m (x) = 0 ha x = 0, de ez nem r´esze az ´ertelmez´esi tartom´anynak x m0 (x) m(x)
7. m00 (x) = 8.
x→1+
0
−2−2x2 (x2 −1)2 ,
] − ∞, −1[ − &
]1, ∞[ + %
] − ∞, −1[ − _
]1, ∞[ − _
m00 (x), seholsem nulla. x m (x) m(x) 00
10. Rm = R (9) Hat´ arozza meg az al´ abbi f¨ uggv´enyek x0 k¨ ur¨ oli n-ed fok´ u Taylor polimonj´at! (a) Tf,0,4 (x) = 3 + x − 2x2 + x3 + x4 Tf,1,4 (x) = 4 + 4(x − 1) + 7(x − 1)2 + 5(x − 1)3 (x − 1)4 = 3 + x − 2x2 + x3 + x4 4 (b) Tf,0,3 (x) = e − 2ex + 2ex2 − ex3 3 4 2 Tf,0,4 (x) = e − 2ex + 2ex2 − ex3 + ex4 3 3 3 2 1 1 4 1 +2 x− x− Tf, 21 ,3 (x) = 1 − 2 x − − 2 2 3 2
2 3 4 4 2 1 1 1 1 Tf, 21 ,4 (x) = 1 − 2 x − − e x− + e x− +2 x− 2 2 3 2 3 2 4 −1 −1 −1 −1 2 3 Tf,1,3 (x) = e − 2e (x − 1) + 2e (x − 1) − e (x − 1) 3 4 2 Tf,1,4 (x) = e−1 − 2e−1 (x − 1) + 2e−1 (x − 1)2 − e−1 (x − 1)3 + e−1 (x − 1)4 3 3 1 1 1 (c) Tf,0,3 (x) = ln 2 + x − x2 + x3 2 8 24 1 1 Tf,−1,3 (x) = x + 1 − (x + 1)2 + (x + 1)3 2 3
VI. feladatsor
(1) Az alapf¨ uggv´enyek integr´ alja seg´ıts´eg´evel hat´arozza meg az al´abbi f¨ uggv´enyek hat´arozatlan integr´alj´at: Z Z 9/2 x x−4 (a) f (x) dx = − ln |x| + 2x + C (b) f (x) dx = 3 ln |x| − 2 +C 9/2 −4 Z Z 1 (c) f (x) dx = ex + 4 cos x + tg x + C (d) f (x) dx = arcsin x + C 2 Z Z 2 5x x6 (e) f (x) dx = arctg x + 4x + C (f) f (x) dx = − 3 + 2 ln |x| + 3x + C 3 ln 5 6 Z Z √ x−1 (g) f (x) dx = 4 sin x + 3ctg x + C (h) f (x) dx = 7 ln |x| + 4 − 3 3x + C −1 (2) Hat´ arozza meg az al´ a bbi f¨ u ggv´ e nyek hat´ a rozatlan integr´ a lj´ a t helyettes´ ıt´ e ssel: Z Z ln |2x + 3| (1 − 7x)3 +C (b) f (x) dx = 3 +C (a) f (x) dx = −3 · 7 2 Z Z e2−4x cos(5 − 3x) (c) f (x) dx = +C (d) f (x) dx = − +C −4 −3 Z Z −ctg (4x + 1) arctg (4x + 3) (e) f (x) dx = + C (f) f (x) dx = 2 +C 4 4 2/3 (5x+3) Z Z arcsin(5x + 2) 2/3 (g) f (x) dx = + C (h) f (x) dx = − +C 5 5 (3) Hat´ arozza meg az al´ abbi f¨ uggv´enyek hat´ arozatlan integr´alj´at helyettes´ıt´essel: Z Z cos6 x 1 (3 + 2x3 )2/3 (a) f (x) dx = −4 +C (b) f (x) dx = +C 6 3 2/3 Z Z 1 tg 4 x (x2 + 2)3/4 +C (d) f (x) dx = +C (c) f (x) dx = 3/4 2 4 Z Z 3 6/5 5/3 1 (x − 2) arctg x (e) f (x) dx = + C (f) f (x) dx = +C 3 6/5 5/3 Z Z 5 ln x 4 (g) f (x) dx = 3 +C (h) f (x) dx = +C 5 arccos x Z Z 1 1 1 1 (i) f (x) dx = − +C (j) f (x) dx = − +C 2 sin2 x 3 (x2 + x)3 (4) Hat´ arozza abbi f¨ uggv´enyek hat´ arozatlan Z meg az al´ Z integr´alj´at helyettes´ıt´essel: 1 7 (a) f (x) dx = ln |4x + 1| + C (b) f (x) dx = − ln |4 − 3x| + C 2 3 Z Z 5 4 (c) f (x) dx = − ln |1 + x2 | + C (d) f (x) dx = ln |1 + 3 sin x| + C 2 3 Z Z 4 (e) f (x) dx = − ln |5 + 3e−x | + C (f) f (x) dx = 2 ln |arctg x| + C 3 (5) Hat´ arozza meg az al´ a bbi f¨ uggv´enyek hat´ arozatlan integr´alj´at helyettes´ıt´essel: Z Z 1 1 1+sin x (a) f (x) dx = e +C (b) f (x) dx = − e x + C 4 Z Z √ 1 1 +C (c) f (x) dx = sin(5 x) + C (d) f (x) dx = 2 cos 5 x2 (6) Hat´ arozza abbi f¨ uggv´enyek hat´ arozatlan integr´alj´at parci´ Z meg az al´ Zalis integr´al´assal: (a) f (x) dx = (2x − 5) sin x + 2 cos x + C (b) f (x) dx = −(x2 − 1) cos x + 2x sin x + 2 cos x + C Z Z e2x e1−2x (c) f (x) dx = (4x + 3) − e2x + C (d) f (x) dx = 4x − e1−2x + C 2 −2 Z Z p 1 2 (e) f (x) dx = − sin(2x)e1−x − cos(2x)e1−x + C (f) f (x) dx = x arcsin x + 1 − x2 + C 53 5 Z Z x x3 x2 2 (g) f (x) dx = − x + x ln x − + − x + C (h) f (x) dx = x ln x − x + C 3 9 2 Z Z 1 (i) f (x) dx = xarctg x − ln |1 + x2 | + C (j) f (x) dx = x2 arctg x − x + arctg x + C 2 Z 2 3 cos(2x + 2)e3x−4 + sin(2x + 2)e3x−4 + C (k) f (x) dx = 13 13 3 3 Z x x2 x x2 2 x2 (l) f (x) dx = ln2 x + + 2x − 2 ln x + + 2x + x3 + + 4x + C 3 2 9 4 27 4 (7) Hat´ arozza meg az al´ abbi f¨ uggv´enyek hat´ arozatlan integr´alj´at:
Z
Z
x 1 +C (a) f (x) dx = arctg 2 2 Z 4 1 arctg x +C (c) f (x) dx = 20 5 Z 1 x+1 (e) f (x) dx = arctg +C 2 2 Z (g) f (x) dx = arcsin(x − 1) + C Z 1 (i) f (x) dx = arcsin(2x + 3) + C 2
(b) Z (d) Z (f) Z (h) Z (j)
1 arctg (3x + C) 3 p 5/3x 2 arctg p +C f (x) dx = 3 5/3 3 x−3 f (x) dx = arctg +C 2 2 1 f (x) dx = arcsin(3x − 1) + C 3 1 f (x) dx = arcsin(3x − 2) + C 3 f (x) dx =
VII. feladatsor (1) Hat´ arozza meg az al´ abbi hat´ arozott integr´ alok ´ert´ek´et: Z 7 Z 2√2 Z −1 √ √ x 1 38 √ dx = − ln 2 (b) (c) dx = 10 − 3 (a) x + 2 dx = x 3 1 + x2 Z 3π Z−2 Z 2−2 e π 1 dx = (f) sin x dx = 2 (d) ln x dx = 1 (e) 2 8 0 1 −4 x + 8x + 20 (2) Hat´ arozza meg a ter¨ uleteket, ha: Z −3 (a) f (x) dx = ln 4 Z−6 1 e2 + 1 f (x) dx = (b) 4 Z0 1 16 f (x) dx = (c) 3 −3 (3) Hat´ arozza meg a k´et g¨ orbe ´ altal k¨ ozrez´ art ter¨ uletet! (a) T = 4, 5 (b) T = 4, 5 1 (c) T = 3 (4) Hat´ arozza meg az f (x) ´es az x tengely ´ altal k¨ozbez´art ter¨ uletet! 125 (a) T = 6 125 (b) T = 6 5 (c) T = − ln 4 4 (5) Hat´ arozza meg abbi f¨ ugv´enyek ´ıvhossz´ at! √ al´ (a) sf = 2 10 π (b) sf = 2 (c) sf = 74 (6) Hat´ arozza meg al´ abbi f¨ ugv´enyek x tengely k¨or¨ uli megforgat´as´aval keletkezett forg´astest t´erfogat´at! 2 (a) V = π 3 (b) V = π ln 3 (c) V = π e2 + 2e − 1, 5 (d) V = π 23 2e + 1 (e) V = π 9 π2 (f) V = 24 (7) Hat´ arozza meg al´ abbi f¨ ugv´enyek x tengely k¨or¨ uli megforgat´as´aval keletkezett forg´astest pal´astj´anak felsz´ın´et! (a) A = 2π π 1453/2 2 (b) A = − 18 3/2 3 (8) Hat´ arozza meg al´ abbi f¨ ugv´enyek x tengely k¨or¨ uli megforgat´as´aval keletkezett forg´astest teljes felsz´ın´et! (a) A = 3π π 1453/2 2 (b) A = 64π + − 18 3/2 3
VIII. feladatsor (1) y 0 = 2 + 3ex 2 − 2x + y = 2 − 2x + 2x + 3ex = 2 + 3ex (2) y 00 − 3y 0 + 2y + 3 − 2x = 3ex + 4e2x − 3(3ex + 2e2x + 1) + 2(3ex + e2x + x) + 3 − 2x = 3ex + 4e2x − 9ex − 6e2x − 3 + 6ex + 2e2x + 2x + 3 − 2x = 0 (3) Oldja meg az al´ abbi differenci´ alegyenleteket (a) y = sin x + 2x + C, yp = sin x + 2x + 5 x3/2 x3 x3/2 x3 + + x + C, yp = + +x+0 (b) y = 3 3/2 3 3/2 (4) Oldja meg az al´ abbi differenci´ alegyenleteket 1 (a) y = x2 + Cx2 ex 2 x2 (b) y = + C e−x 2 1 ln(x2 + 1) + C (x2 + 1) (c) y = 2 2 1 (d) y = − + Cex 2 (5) Oldja meg √ az al´ abbi differenci´ √ alegyenleteket 2 + 1, y = 3 x2 + 1 (a) y = C x p √ (b) y = Ce2 x (c) y = −1+ Cx 2 (d) y = ln − 2x e +C 3 3 (e) y = C(2x + 1) 2 , yp = 2(2x + 1) 2 (f) y = p tg (x + C) − x (g) y = pC(1 + x2 ) − 1 (h) y = C(x2 − 2) − 1 (6) Oldja meg az al´ abbi differenci´ alegyenleteket (a) y = C1 e−x + C2 e−3x (b) y = C1 e3x + C2 xe3x (c) y = e−2x (C1 sin(x) + C2 cos(x)) −4x (d) y = C1 + C√ 2e √ (e) y = C1 sin( 5x) + C2 cos( 5x) (f) y = C1 e2x + C2 e−2x (g) y = e3x (C1 sin(5x) + C2 cos(5x)) + 5 cos(2x) − 2 sin(2x) + e−4x x2 5x 19 (h) y = C1 e−2x + C2 e−3x + − + 6 18 108 2 1 2 1 (i) yh = C1 e−x + C2 e−3x , yp = x + , y = yh + yp = C1 e−x + C2 e−3x + x + 3 9 3 9 (j) y = C1 e5x + C2 e( − 2x) + 2 cos(2x) + 3 sin(3x) (k) y = C1 e4x + C2 e−2x + x2 − 2x − 8 (l) y = C1 e−x + C2 xe−x + sin(3x) (m) y = C1 e5x + C2 xe5x + 3e3x − e−x (n) y = e−x (C1 sin(3x) + C2 cos(3x)) +x+1 ! !! √ √ 23x 23x −x/2 (o) y = e C1 sin + C2 cos + e2x + x2 2 2 (7) Oldja meg az al´ abbi differenci´ alegyenleteket Rez e2x 2x (a) y = C1 e + C2 + (6x − 3) 2 −4x e (b) y = C1 e−4x + C2 − (12x2 + 16x + 4) 4 (c) y = C1 e2x + C2 xe2x + 5x2 e2x (d) y = C1 sin(3x) + C2 cos(3x) + x(3 + 2 cos(3x))
