Differensial test 2.0

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

dy = f(ax + by + c)dx tenglama qanday almashtirish yordamida o'zgaruvchilari ajralgan tenglama keladi?

• y = zx

• y = zxⁿ

• z = ax + by + c

• y = eᵃˣ

O'zgaruvchilari ajralgan tenglamani belgilang

• m(x,y)dx + n(y)xdy = 0

• dy = f(ax + by)dx

• m(x)dx + n(y)dy = 0

• a(x)y¹ + b(x)y = c(x)

yⁿ + 9y¹ = 0 tenglamaning umumiy yechimini toping:

• y = c₁ + c₂e⁹ˣ

• y = c₁e³ˣ + c₂e ⁻³ˣ

• y=c₁cos3x + c₂sin 3x

• y = c₁ + c₂e⁻⁹ˣ

3x² ydx + (x³ydx + (x³ - 3y²)dy = 0 teglamaning umumiy yechimini toping:

• xy² + y = c

• x²y² - y = c

• xy - y² = c

• x³y - y³ = c

y¹ + 2y = eˣ tenglamaning yechimini toping:

• 
• 
• 
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Hususiy yechimi y₁ = xeˣ bo'lgan chiziqli o'zgarmas koeffisientli differensial tenglamani ko'rsating

• yⁿ + 2yⁿ + y = 0

• yⁿ - 2y¹ - 3y = 0

• y¹ - y = 0

• yⁿ - 2y¹ + y = 0

yⁿ + 4y = 0 tenglamaning umumiy yechimini toping

• y = c₁cos2x + c₂xcos2x

• y = c₁e²ˣ + c₂e⁻²ˣ

• y = ce⁻⁴ˣ

• y = c₁cos2x + c₂sin 2x

• y = ceˣ

• y = c + x

• y = c₁x + c₂

• y = cx

Ushbu y = ce²ˣ chiziqlar sinfning differensial tenglamasini toping

• y¹x² = y²

• 
• y x =3y

• y¹ = 2y