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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

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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

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

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EllisMaster · Tutor for 5 years

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<p> 4.1.2/5.3</p>

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<p> The question is asking us to arrange the given ordinary differential equations with constant coefficients in ascending order according their order.<br /><br />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.<br /><br />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. <br /><br />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.<br /><br />4. With uy′+u2uy-u5=u0 equation, the order is 1. This is based on the highest derivative present. <br /><br />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.<br /><br />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.<br /><br />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).</p>
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