Bai5 PTVPC2 HK2 0506 Web(1)

Bai5 PTVPC2 HK2 0506 Web(1) - BO M ON TOAN NG DU NG HBK...

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BOÄ MOÂN TOAÙN ÖÙNG DUÏNG - ÑHBK ------------------------------------------------------------------------------------- TOAÙN 4 CHUOÃI VAØ PHÖÔNG TRÌNH VI PHAÂN BAØI 5: PHÖÔNG TRÌNH VI PHAÂN CAÁP 2 TS. NGUYEÃN QUOÁC LAÂN (5/2006)
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NOÄI DUNG ----------------------------------------------------------------------------------------------------------------------------------- 1 – PHÖÔNG TRÌNH VI PHAÂN CAÁP 2. TRÖÔØNG HÔÏP GIAÛM CAÁP 3 – PHÖÔNG TRÌNH VI PHAÂN CAÁP 2 HEÄ SOÁ HAØM 2 – PHÖÔNG TRÌNH VI PHAÂN CAÁP 2 HEÄ SOÁ HAÈNG
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GIAÛM CAÁP PHÖÔNG TRÌNH VI PHAÂN CAÁP 2 -------------------------------------------------------------------------------------------------------------------------------------- VD: Giaûi phöông trình vi phaân caáp 2: x x x y y cos ' ' ' + = Phöông trình vi phaân caáp 2: F(x, y, y’, y’’) = 0 BT Coâsi: PT chuaån hoaù + ÑK ñaàu ( 29 ( 29 ( 29 = = = 1 0 0 0 ' , ' , , ' ' y x y y x y y y x f y Giaûm caáp cô baûn: Phöông trình F(x, y’, y’’) = 0 Nguyeân taéc: Ñaët u(x) = ñaïo haøm caáp thaáp nhaát cuûa aån y ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 0 ' , , : " ' ' 0 ' ' , ' , = = = = u u x F x y x u x y x u y y x F Nghieäm toång quaùt PT vi phaân caáp 2 chöùa 2 haèng soá C 1 , C 2 Ñaùp soá: Nghieäm x x x C x C y cos sin 2 2 1 - + + =
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PHÖÔNG TRÌNH VI PHAÂN TUYEÁN TÍNH CAÁP 2 -------------------------------------------------------------------------------------------------------------------------------------- Tuyeán tính (linear): y,y’,y’’ – baäc 1 Heä soá haøm, k0 thuaàn nhaát (veá phaûi): y’’ + p(x)y’ + q(x)y = f(x) duï: Heä soá haèng, k0 thuaàn nhaát (coù veá phaûi): y’’ + py’ + qy = f(x) duï: ( 29 1 sin cos sin ' ' ' 2 x x x y e x y xy x + = - + ( 29 3 sin cos 4 ' 3 ' ' x x x y y y + = - + PT thuaàn nhaát töông öùng: y’’ + p(x)y’ + q(x)y = 0 ( 29 2 0 sin ' ' ' 2 = - + y e x y xy x Ví duï: Töông öùng (1): PT thuaàn nhaát töông öùng: y’’ + py’ + qy = 0 Ví duï: Töông öùng (3): ( 29 4 0 4 ' 3 ' ' = - + y y y
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GIAÛI PTVP TUYEÁN TÍNH C2 THUAÀN NHAÁT HEÄ SOÁ HAÈNG ---------------------------------------------------------------------------------------------------------------------------------------------- x k x k tn tq e C e C y 2 1 2 1 . + = x k x k e e 2 1 , sôû nghieäm 2 PTVPC2 thuaàn nhaát heä soá haèng y’’ + py’ + qy = 0 PTrình ñaëc tröng k 2 + pk + q = 0 > 0: k 1 k 2 R < 0: N0 phöùc = 0: k 1 = k 2 R β α i k i m ± = = < 2 , 1 2 2 0 kx kx xe e , sôû nghieäm 2 kx kx tn tq xe C e C y 2 1 . + = (thöïc sôû nghieäm 2 [ ] x C x C e y x β β α sin cos 2 1 + = x e x e x x β β α α sin , cos Phaûi tìm ñuû 2 nghieäm phöông trình ñaëc tröng
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y’’’ –y = 0 y’’ – 5y’ + 6y = 0 y’’ – 4y’ + 4y = 0 y’’ – 2y’ + 5y = 0
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