ответА) ав(5в-2)
Пошаговое объяснение:
2а(2в²-в)+ав²=4ав²-2ав+ав²=5ав²-2ав=ав(5в-2)
My recent winter holidays were a lot of fun. I tried to visit as many places as I could and to do a lot of different things. Even though the weather wasn't very nice, it didn't make me feel sad. I went to the mall and to the skating-rink with my family and friends. It's always interesting to learn something new, so I also visited the movies and watched some great films. And then, of course, there were New Year and Christmas, which are both my favourite holidays. We had a gorgeous Christmas-tree decorated with toys and Christmas lights. On the first day of new year I got many presents from my family and relatives. After all, it was just great to take some time off from school and to sleep till noon. I really enjoyed my winter holidays.
Его характеристическое уравнение имеет вид:
k² + 4 = 0
k² = -4
Его корни k₁,₂ = 2i.
То есть в данном случае корни комплексные(k₁=α+βi,k₂=α-βi) и для них α = 0,β =2 Следовательно, решение однородного уравнения запишется в виде:
y(x) = C₁cos(βx) +C₂sin(βx) = C₁cos(2x) +C₂sin(2x)
Для нахождения функций C₁ и C₂ используем начальные условия:
y(0)=1; y'(0) = 2
y(0) =C₁cos(2*0) + C₂sin(2*0) = C₁ = 1.
Найдем производную функции:
y'(x) = -2C₁sin(2x) + 2C₂cos(2x).
Подставим начальное условие:
y'(0) = -2sin(0) + 2C₁cos(0) = 2С₁ = 2 ⇒С₁ = 1.
Следовательно частное решение дифференциального уравнения:
y(x) = cos(2x) + sin(2x)
Проверка: y'(x) = -2sin(2x) + 2cos(2x)
y''(x) = -4cos(2x) - 4sin(2x)
Подставляем в исходное уравнение
y'' + 4y = -4cos(2x) - 4sin(2x) + 4(cos(2x)+sin(2x)) = 0
ответ: y(x) = cos(2x) + sin(2x)
Привет!
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