I-bob. Birinchi tartibli differensial tenglamalar
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= yechimlarni olamiz, bundan 1 2
( ) 3 2 p u p
p p C
′ = = − + , ya’ni 1 2
3 2 dp dt p p C = − + . Oxirgi tenglikni integrallab, 1 2 ( ) ( ,
, ) p t t C C = Φ
ni topamiz, bu yerda Φ -ma’lum funksiya. Toplilgan ( ) p t
funksiyani ( )
t z e p t − = ga qo’yib, bu ni esa ( )
y z x
′ = almashtirishga qo’yamiz va uni bir marta integrallab berigan tenglama yechimini 3 1 2 3 ( ) ( , , ) t t y e p t dt C
e t C C dt C t x
− − = + = Φ + =
ko’rinishda topamiz. 5- Masala. 2 2 2 6 3 4 y x y
xy y x ′′ ′ − = −
tenglamaning (1) 1,
(1) 4 y y′ = = shartlarni qanoatlantiruvchi xususiy yechimini toping. Yechish. Berilgan tenglamada 1 2 , , , m m m x tx z
t z z t z z t z − − ′ ′ ′′ ′′ = = = =
almashtirishlarni bajarib, umunlashgan bir jinsli tenglama bo’lishini abiqlaymiz va 2 m
da ya’ni, 2 , ( ) t t x e y e z t = = almashtirishlarda berilgan tenglama 2 6
z z ′′ − = tenglamaga keladi. Oxirgi tenglamani ikkala tomonini 2z′
ko’paytirib, 2 2 12 0 z z z z ′ ′′
′ − = ya’ni ( ) ( )
2 3 4 0 z z ′ ′ ′ − =
ni hosil qilamiz endi esa bu tenglikni integrallab, 2 3 4 1 z z C ′ = +
topamiz. Masala qo’yilishidagi (1) 1,
(1) 4 y y′ = = sartlardan, hamda , t x e = 2 ) ( t y e z t = va ( 2 )
t y e z z ′ ′ = + almashtirishlarga asosan (0) 1 z = va (0) 2 (0) 4 z
′ + = ya’ni (0)
2 z′ = shartlarga ega bo’lamiz. Demak 78
1 2 3 (0) 4 (0) z z C ′ = + , ya’ni 1 0 C = . Shunday qilib, 2 3 4 z z ′ = yoki 3 2 2 z z ′ = ±
tenglamani integrallash orqali quyidagini topamiz 2 1 t C z ± = + , bundan esa (0) 1 z
shartga asosan 2 1 C = ±
, ya’ni 2 1 ( 1) z t = ± . Bu yechimlardan (0) 2
= shartni qanoatlantiruvchi 2 1
1) z t = − yechimni topamiz. Shunday
qilib, berilgan tenglama yechimi 2 2
2 2 ( ) ( 1) (ln 1) t e x t y e z t t x = = = − − bo’ladi. Mustaqil yechish uchun mashqlar: I. Quyidagi differensial tenglamalarni integrallang (288-307):
288. ln y xy y x ′ ′′ ′ = .
289. 2 2 1 xy y
y ′ ′′
′ = + .
290. 2 ( ) x a y xy y ′′ ′ ′ + + = . 291.
4 3 2 1 x y
x y ′′′
′′ + = .
292. 2 4 4 y y xy ′ ′′ ′′ + =
293.
2( 1) y y ctgx
′′′ ′′ = − .
294. 2 2 y y y y x ′ ′′ ′ ′′′ − = .
295. 3 0 y xy y ′′ ′′′ ′′′
− + = .
296. 2 1 ( 1) 2 2 x x y y x + ′′′ ′′ − + = . 297. 3 1 y y ′′ = .
298. 2 2 1 y y ′′ = .
299. 2 2 yy y y y ′′ ′ ′ − = . 300.
3 1 IV y y y′′
− = . 301.
2 2 1 0 y y y
′′ ′ ′′′
− + =
.
302. sin y xy y x x ′ ′′ ′ = + . 303.
2 2 ln yy yy y y ′′ ′ ′ − =
304. 2 , (2) 0, (2)
4 y x y y y x y ′ ′′ ′ = + = = ′ . 305.
2 , (0) 0, (0) 1
y y e y y ′′ ′ = = = ,
306. 2 2 3 0; (0) 3, (0) 1,
(0) 1 y y y y y ′′′
′ ′ ′′ − = = − = = −
.
307. 2 cos
sin ; ( 1) , ( 1)
2 6 y y y y y y y π ′′ ′ ′ ′ + = − =
− = .
79
II. Quyidagi differensial tenglamalarni (bir jinsli ekanligidan foydalanib) integrallang (308-324): 308. 2
xyy xy yy ′′ ′ ′ − − = .
309. 2 2 ( ) x yy y xy ′′ ′ = − . 310.
2 2 2 2 2 ( ) x yy y xyy
y x y y ′′ ′ ′ ′ − + = − . 311.
2 2 xyy xy yy ′′ ′ ′ + = . 312. 2 2 1 yy yy y x ′ ′′ ′ − = + .
313.
2 2 y y y y x y x ′ ′ ′′ + + = . 314. 2 2 2 ( 2 ) x y yy y ′ ′′ − = .
315. 2 ( ) (1 ) y xy y xy x ′′ ′ ′ + = −
316. 2 2 2 2 0 bxy xyy xy yy a x ′ ′′ ′ ′ − − − = − 317. 2 3 2 4 4x y y x y ′′ = − . 318. 2 3 0 xyy
yy x y
′′ ′ ′ + − = .
319. 2 2 3 4 0 x y xy y x ′′ ′ − + + = 320.
2 2 3 ( ) (2 3 ) x yy y xyy
xy y x ′ ′ ′ ′ − + = − . 321. 4 3 ( ) 0 x y xy y ′′ ′ + − = . 322. 4 2 3 ( 2 ) 4 1 x y yy x yy ′ ′′ ′ − = + .
323. 2 3 yy xyy xy x ′ ′′ ′ + − = . 324.
4 3 3
2 2 2 3 2 3 2 (3 2 ) 2 0 x y x y
x yy xy x y x y
y ′′ ′ ′ ′ − + − + + +
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