Question

Y''-8y'+16y=0 решение линейного дифференциального уравнения подробно

Answer 1

Пусть $y=e^{kx}$, получим характеристическое уравнение

$k^2-8k+16=0\\ (k-4)^2=0\\ k=4$

Общее решение: $y=C_1e^{4x}+C_2xe^{4x}=e^{4x}(C_1+C_2x)$

Answer 2

#### Solving the Linear Differential Equation Y'' - 8Y' + 16Y = 0 To solve the linear differential equation Y'' - 8Y' + 16Y = 0, we can use the method of characteristic equation. This method involves assuming a solution of the form Y = e^(rt), where r is a constant to be determined. Let's proceed with the solution step by step: Step 1: Find the characteristic equation. The characteristic equation is obtained by substituting Y = e^(rt) into the differential equation and simplifying. In this case, the differential equation is Y'' - 8Y' + 16Y = 0. Substituting Y = e^(rt) into the differential equation, we get: (e^(rt))'' - 8(e^(rt))' + 16(e^(rt)) = 0 Step 2: Differentiate Y. Differentiating Y = e^(rt) twice, we get: Y' = re^(rt) Y'' = r^2e^(rt) Step 3: Substitute the differentiated Y into the differential equation. Substituting Y = e^(rt), Y' = re^(rt), and Y'' = r^2e^(rt) into the differential equation, we get: r^2e^(rt) - 8re^(rt) + 16e^(rt) = 0 Step 4: Simplify the equation. Factoring out e^(rt), we have: e^(rt)(r^2 - 8r + 16) = 0 Step 5: Solve the characteristic equation. Setting the expression in the parentheses equal to zero, we get: r^2 - 8r + 16 = 0 To solve this quadratic equation, we can use the quadratic formula: r = (-b ± √(b^2 - 4ac)) / (2a) In this case, a = 1, b = -8, and c = 16. Substituting these values into the quadratic formula, we get: r = (-(-8) ± √((-8)^2 - 4(1)(16))) / (2(1)) r = (8 ± √(64 - 64)) / 2 r = (8 ± √0) / 2 r = 8 / 2 r = 4 Step 6: Determine the solution. Since we obtained a repeated root of r = 4, the general solution of the differential equation is: Y = C1e^(4t) + C2te^(4t) where C1 and C2 are arbitrary constants. This is the detailed solution to the linear differential equation Y'' - 8Y' + 16Y = 0. If you have any further questions, feel free to ask!