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WorksheetsDifferensial test 1.0
Total questions: 21
Worksheet time: 32mins
Berilgan tenglamaning tipini aniqlang:
(xcos y + y)dy + (sin y + x)dx = 0
To'la differensial
Bernulli
Bir jinsli differensial tenglama
y ga nisbatan chiziqli
Integrallovchi kopaytuvchi qanday tenglamalar uchun kiritiladi ?
Chiziqli va to'la differensial
Bernulli
Bir jinsli differensial tenglama
Rikkati
Berilgan tenglamani tipini aniqlang:
Bernulli
Chiziqli
Bir jinsli differensial tenglama
Rikkati
Ushbu masala qanday masala deyiladi
Koshi masalasi
Chegaraviy masala
Variatsion masala
To'g'ri javob yo'q
Berilgan tenglamani nomini aniqlang
Chiziqli
Bernulli
O'zgaruvchilari ajraladigan
Rikkati
Chiziqli erkli funksiyalar sistemasini aniqlang
x , x², x³
eˣ , 2eˣ , 3e ˣ
x - 1, 2x - 2
Bunday funksiya yo'q
Quyidagi funksiyalar sistemasining qaysi biri chiziqli bog'liq?
sin2x, cos x
1, x, x²
x, 3x, 6x
eˣ , e²ˣ, e³ˣ
Quyidagi tenglamalarning qaysi biri Bernulli tenglamasi deyiladi:
m(x)ydx + n (y)xdy = 0
dy = f ( ax + by )dx
a(y)x¹ + b(y)x = c(y)xⁿ
mx(x)dx + n(y) dy = 0
Birinchi tartibli chiziqli differensial tenglama qaysi javobda to'gri ko'rsatilgan ?
m(xy)dx + n(y)xdy = 0
dy = f (ax + by)dx
(y)x¹ + b(y)x = c(y)
m(x)dx + n(y)dy = 0
Agar Rikkati tenglamasida y₁ xususiy yechimi ma'lum bo'lsa,
bu tenglama qanday almashtirish yordamida
yechiladi?
y₁ = zx
y₁ = zy
y = z + y₁
y = eᶻʸ¹
F (x,y,y¹,yⁿ,....,y(ⁿ) = 0 tenglama y va uning hosilalariga nisbatan
birjinsli bo'lsa, nu tenglama tartibi ... almashtirish orqali bittaga kamaytiriladi
y¹ = zx
y¹ = z
y¹ = zy
y¹ = zeˣ
Hosilaga nisbatan yechilmagan tenglamani ko'rsating
a(x)y¹ + b(x)y = c(x)yᵃ
m(x)dx + n(y)dy = 0
y = xy¹ + ᵩ(y¹)
(x)y¹ + b(x)y + c(x)yᵃ = d(x)
y¹ = f(x,y) tenglama uchun Koshi shartini aniqlang
y¹(x₀) = y₀
x₀ = 1
y(x₀) = y₀
y₀ = 0
Xususiy yechimlari eˣ , shx, chx bo'lgan eng kichik tartibli chiziqli differensial tenglama tuzing
yⁿ + y = 0
yᵐ - y¹ = 0
yⁿ - y = 0
yᵐ + 3yⁿ = 0
Quyidagi funksiyalar sistemasining qaysi biri chiziqli bog'liq?
sin x, cos x
1, x, x²
6x + 9, 8x + 12
eˣ , e²ˣ, e³ˣ
Lagranj tenglamasini aniqlang
a(x)y¹ + b(x)y = c(x)yᵃ
m(x)dx + n(y)dy = 0
y = xф(y¹) + ᵠ(y¹)
a(x)y¹ + b(x)y + c(x)yᵃ = d(x)
M (x,y)dx + N(x,y)dy = 0 tenglama to'la differensial tenglama bo'ladi, agar tenglik o'rinli bo'lsa
M(x,y)dx + N(x,y)dy = 0 tenglama o'zgaruvchilarga nisbatan bir jinsli deyiladi,
agar .... tenglik o'rinli bo'lsa
M (tx,ty) = tⁿM (x,y),
N(tx,ty) = tᵏN= N (x,y), k = n
M(tx,ty) = tⁿN(x,y)
M(tx,ty) = tⁿM(x,y) ,
N (tx,ty) = tⁿ N(x,y)
M (tx,ty) = tⁿ M (x,y) ,
N (x,y) ixtiyoriy
Quyidagi tenglamalarning qaysi biri Rikkati tenglamasi deyiladi:
m(x)ydx + n(y)xdy = 0
dy = f(ax + by)dx
a(x)y¹ + b(x)y + c(x)y² = d(x)
m (x)dx + n(y)dy = 0
Bernulli tenglamasini belgilang:
m(x) ydx + n (y)xdy = 0
dy = f (ax + by)dx
x¹ + b(y)x = c(y) xⁿ
m(x)dx + n(y)dy = 0
Birinchi tartibli chiziqli tenglamani aniqlang:
m(x)ydx + n(y)xdy = 0
dy = f (ax + by)dx
a(x)y¹ + b(x)y = c(x)
m(x)dx + n(y)dy = 0
