Question
Обыкновенные линейные дифференциальные уравнения с постоянными коэффициентами расположить по возрастанию порядков этих уравнений: ВАРИАНТЫ ОТВЕТОВ y^(''')+5y^(')=0 y^((6))+2y^(')=0 y^((4))+3y^((2))-7=0 y^(')+2y-5=0 y^('')+3y-2=0
Answer
4.7
(392 Votes)
Ellis
Master · Tutor for 5 years
Answer
4.1.2/5.3
Explanation
The question is asking us to arrange the given ordinary differential equations with constant coefficients in ascending order according their order.1. The differential equation y^′′+5y′=0 is a second order differential equation. This is because the order of a differential equation is determined by the order of the highest derivative, herein it is 2nd implying it is a second-order differential equation.2. The shown equation uy^(6)+u2uy′=u0 has the highest order of differential equation as 6 proving it to be a sixth order differential equation. 3. The given equation uy^(4)+3uy^(2)-7=0 is a standard algebraic equation and not a differential equation due to the absence of derivatives. Therefore, it does not feature any order.4. With uy′+u2uy-u5=u0 equation, the order is 1. This is based on the highest derivative present. 5. In case of uy^′′+3uy-u2=u0, the order of the equation is 2 seeing that the highest order derivative is the second one.By examining all equations, their order goes in the decreasing sequence as 6, 2, 1. However, the third statement is deemed irrelevant as it lacks any differential form with order.Ordered as per the question, the equations as per ascendant order are: uy′+ u2uy - u5 = u0 (Equ. with actual order of 1); y^′′+5y′=0 and uy^′′+3uy - u2=u0 (Equations each with an order of 2); uy^(6) + u2uy' = u0 (An equation of the order at 6) ; uy^(4) + 3uy^(2) - 7 = 0 (an equation with no existing order).