Ixtiyoriy natural son bo’lsa, unda mavhum birlik

Ta’rifni to‘ldiring: Differensial tenglamaning tartibi deb unda qatnshuvchi noma’lum funksiya hosilalarning ……… aytiladi

eng katta tartibiga

eng katta darajasiga

eng katta qiymatiga

Soniga

to‘g‘ri javob keltirilmagan

(y′)^3–(y′)^2+ y′′–y+5y^4+x^5=0 differensial tenglama nechanchi tartibli ?

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Differensial tenglamaning yechimi yana nima deb ataladi ?

integral .

tenglashtiruvchi funksiya .

boshlang’ich funksiya .

differensial .

ildiz .

Quyidagi funksiyalardan qaysi biri y′−2y=0 differensial tenglamaning yechimi bo’ladi?

y=e^2^x .

y=ln2x .

y=cos2x .

y=sin2x .

y=x^2 .

n-tartibli F(x, y, y′, ..., y^(^n−^1), y^(^n^))=0 differensial tenglama uchun boshlang‘ich shartlar qanday ko‘rinishda bo‘ladi

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I tartibli differensial tenglama eng umumiy holda qanday ko‘rinishda bo‘ladi ?

F(x,y,y′)=0 .

F(x,y)= y′ .

F(x, y′)= y .

F(x,y,y′, y′′)=0 .

F(y,y′)= x .

I tartibli differensial tenglama uchun boshlang‘ich shart qanday ko‘inishda bo‘ladi ?

y(x0)= y0 .

y′(x0)= y0 .

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I tartibli differensial tenglama uchun Koshi masalasini ko‘rsating

y′=f(x,y) , y(x0)= y0 .

y′=f(x,y) , y′(x0)= y0 .

y′=f(x0,y) .

y′=f(x,y0) .

y′=f(x0,y0) .

I tartibli o‘zgaruvchilari ajralgan differensial tenglamani ko‘rsating

M(x)dx+N(y)dy=0 .

M1(x) N1(y)dx+ M2(x)N2(y)dy=0 .

y′+P(x)y=Q(x) .

y′=f(x/y) .

y′=f(xy) .

O‘zgaruvchilari ajralgan differensial tenglamaning umumiy integralini toping.

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I tartibli o‘zgaruvchilari ajraladigan differensial tenglamani ko‘rsating

M1(x) N1(y)dx+ M2(x)N2(y)dy=0 .

M(x)dx+N(y)dy=0 .

y′=f(xy) .

y′=f(x/y) .

y′+P(x)y=Q(x) .

O‘zgaruvchilari ajraladigan differensial tenglamaning umumiy yechimini toping

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I tartibli bir jinsli differensial tenglamani ko‘rsating .

y′=f(x/y) .

y′=f(xy) .

y′+P(x)y=Q(x) .

M1(x) N1(y)dx+ M2(x)N2(y)dy=0 .

M(x)dx+N(y)dy=0 .

I tartibli bir jinsli differensial tenglamani qanday almashtirma yordamida integrallanadi

y=ux .

y=x/u .

y=u–x .

y=u+x .

y=u/x .

I tartibli chiziqli differensial tenglamani integrallang

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Bernulli tenglamasi qaysi javobda to’g’ri ifodalangan?

y′+P(x)y =Q(x)y^n (n≠0, n≠1 ) .

y′+P(x)/y^n=Q(x) (n≠0, n≠1 ) .

y′+P(x)y^n=Q(x) (n≠0, n≠1 ) .

y′+P(x)y =Q(x)/y^n (n≠0, n≠1 ) .

y^n y′+P(x)y=Q(x) (n≠0, n≠1 ) .

y′+P(x)y =Q(x)y^n (n≠0, n≠1 ) Bernulli tenglamasi qanday almashtirma yordamida chiziqli tenglamaga keltiriladi

z=y^1^−n .

z=y^n+^1

z=y^n−^1 .

z^n=y .

z=y^n .

y′+xy =x^3y^3 Bernulli tenglamasining umumiy yechimini topish uchun qanday almashtirma bajariladi

z=y^−^2 .

z=y^−^3 .

z=y^2 .

z^3=y .

z=y^3 .

Qaysi shartda M(x,y)dx+N(x,y)dy=0 to‘liq differensialli tenglama bo‘ladi

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Quyidagilardan qaysi biri to‘liq differensialli tenglama bo‘ladi

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tenglama α parametrning qanday qiymatida to’liq differensialli tenglama bo’ladi?

α=1 .

α=0 .

α=−1 .

α=±1 .

α≠±1 .

II tartibli differensial tenglamaning umumiy ko‘rinishi qayerda to‘g‘ri ko‘rsatilgan

F(x, y, y′, y′′)=0 .

F(x0, y0, y′, y′′)=0 .

F( y, y′, y′′)=0 .

F(x, y, y′′)=0 .

F(x, y′′)=0 .

II tartibli hosilaga nisbatan yechilgan differensial tenglamaning umumiy ko‘rinishi qayerda to‘g‘ri ko‘rsatilgan

y′′= f(x, y, y′) .

y′′= f(x0 , y0, y′) .

y′′= f(y, y′) .

y′′= f( y, y′, y′′) .

y′′= f(x, y, y′, y′′) .

II tartibli differensial tenglama uchun Koshi masalasi qayerda to‘g‘ti ifodalangan

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Quyidagi II tartibli differensial tenglamalardan qaysi birini tartibni pasaytirish usulida integrallab bo‘lmaydi

y′′= xy .

y′′= xy′ .

y′′= yy′ .

y′′= x .

barcha tenglamalarni tartibni pasaytirish usulida integrallab bo‘ladi .

y′′=sin2x differensial tenglamaning umumiy yechimini toping

y= – 0.25sin2x+ C1x+ C2 .

y= – 0.25cos2x+C .

y= – 0.25sin2x+C .

y= – 0.25cos2x+C1x+ C2 .

to‘g‘ri javob keltirilmagan .

y′′=f(x,y′) ko‘rinishdagi differensial tenglamani tartibni pasaytirish usulida integrallash uchun qanday almashtirmadan foydalaniladi ?

y′= p(x) .

y′′= p(x) .

y=p(x) .

y′= p(y) .

y′′= p(y) .

y′′=y′ tenglamaning umumiy yechimini tartibni pasaytirish usulida toping

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y′′=y′/x tenglamaning umumiy yechimini tartibni pasaytirish usulida toping

y=C1+C2x^2.

y=(C1+C2 x)/x^2 .

y=C1+C2x .

y=C1+C2/x .

y=C1+C2/x^2 .

y′′=f(y,y′) ko‘rinishdagi differensial tenglamani tartibni pasaytirish usulida integrallash uchun qanday almashtirmadan foydalaniladi

y′=p(y) .

y′′=p(y) .

y′′= p(x) .

y′= p(x) .

y=p(x) .