o’zgaruvchilari ajraladigan tenglamalar. birinchi tartibli tenglamalar

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1662970369.doc dy dx f x g y = ( ) ( ) ¹ dx x f y g dy ) ( ) ( = ò ò + = c dx x f y g dy ) ( ) ( x y ò ò = + - = + - = . , ln ln ln , x c y c x y yani c x dx y dy dy dx f x y = ( , ) f y x f x y ( , ) ( , ) 1 = ) ( ) , 1 ( 1 х у x y f y j = = u u dx du x u u x u - = = + ¢ ) ( ) ( j j ò ò ò = - + = - = - cx u u du c x dx u u du x dx u …

a ko’paytirib va g(y) 0 ga bo’lib tenglikni hosil qilamiz. uni integrallab yechimni topish mumkin: embed equation.3 misol. dy/dx=- tenglama yechilsin. yechish. o’zgaruvchilarni ajratib dy/y=-dx/x integrallaymiz: 1 – ta’rif. agar ( ning har qanday qiymatida f((x,(y)=(kf(x,y) tenglik bajarilsa, f(x,y) funksiya x va y o’zgaruvchilarga nisbatan k - tartibli bir jinsli funksiya deyiladi. 2 – ta’rif. agar birinchi tartibli (1.2) differensial tenglamaning o’ng tomoni - f(x,y) 0-tartibli bir jinsli funksiya bo’lsa, u holda (1.2) tenglama bir jinsli tenglama deyiladi. f(x,y) nolinchi tartibli bir jinsli bo’lsa, u holda ixtiyoriy ( uchun f((x,(y)=f(x,y) bo’ladi. xususan, u holda (1.2) tenglama (2) bu tenglamani yechish uchun y/x=u deb olamiz. u holda y=ux, y’=u’x+u. bularni (2) ga qo’yib o’zgaruvchilari ajraladigan tenglamaga kelamiz. integrallagandan so’ng u ni o’rniga u/x ni qo’ysak, (2) tenglamaning umumiy integrali hosil bo’ladi. misol. - 0-tartibli bir jinsli funksiya. tenglamani quyidagicha yozib olamiz: uzgaruvchilari ajraladigan differentsial tenglamani hosil qilamiz. natijada 3–ta’rif. birinchi tartibli …

y1-n ifodani qo’yib, bernulli tenglamasining umumiy yechimini hosil qilamiz. a d a b i yo t l a r 1. a.s. piskunov. differensial va integral hisob. t. «o’qituvchi»,1974 y ,17 – 31 betlar. l.e.elsgolts. differensialnie uravneniya i variatsionnoe ischislenie. m. ,»nauka» , 1969 g. ,s . 24 – 31 . 3. l.s. pontryagin. differensialnie uravneniya i ix prilojeniya. m., nauka , 1965 g., s.15 – 20. 4. m.s. salohitdinov, o’.n. nasritdinov. oddiy differensial tenglamalar. t. «uzbekiston» , 1994 y., 19 – 31 betlar . 5. v.p. minorskiy. oliy matematikadan masalalar to’plami. t. «o’qituvchi», 1977, 224-230 betlar. 6. www.ziyonet.uz _1445726621.unknown _1445726629.unknown _1445726633.unknown _1445726637.unknown _1445726639.unknown _1445726641.unknown _1445726642.unknown _1445726640.unknown _1445726638.unknown _1445726635.unknown _1445726636.unknown _1445726634.unknown _1445726631.unknown _1445726632.unknown _1445726630.unknown _1445726625.unknown _1445726627.unknown _1445726628.unknown _1445726626.unknown _1445726623.unknown _1445726624.unknown _1445726622.unknown _1445726617.unknown _1445726619.unknown _1445726620.unknown _1445726618.unknown _1445726615.unknown _1445726616.unknown _1445726614.unknown

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1662970369.doc dy dx f x g y = ( ) ( ) ¹ dx x f y g dy ) ( ) ( = ò ò + = c dx x f y g dy ) ( ) ( x y ò ò = + - = + - = . , ln ln ln , x c y c x y yani c x dx y dy dy dx f x y = ( , ) f y x f x y ( , ) ( , ) 1 = ) ( ) , 1 ( 1 х у x y f y j = = u u dx du x u u x u - = = + …

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