Question

У''-4у'+4у=0 как решить

Answer 1

У нас уравнение вида y''+py'+qy    p=-4   q=4

такое уравнение - линейное  второго порядка с постоянными коэффициентами.  Решим это однородное у-е, для чего составим характеристическое ур-е
k^2-4k+4=0    ⇒ (k-2)^2=0    один корень сдвоенный k1=2
Решение дифф. уравнения имеет вид  y(x)=C1*e^(k1*x) +C2*x*e^(k1*x) =
=C1*e^(2x)+C2*x*e^(2x)   это ответ, можно записать в виде
=e^(2x)*(C1+C2*x)

Answer 2

Solving the Differential Equation

To solve the differential equation у'' - 4у' + 4у = 0, we can use the method of characteristic equations. Let's go through the steps:

1. Assume a solution of the form у = e^(rt), where r is a constant to be determined. 2. Take the first and second derivatives of у with respect to t. - The first derivative is у' = re^(rt). - The second derivative is у'' = r^2e^(rt). 3. Substitute the derivatives back into the differential equation: r^2e^(rt) - 4re^(rt) + 4e^(rt) = 0. 4. Factor out e^(rt): e^(rt)(r^2 - 4r + 4) = 0. 5. Since e^(rt) is never zero, we can divide both sides of the equation by e^(rt): r^2 - 4r + 4 = 0.

Now, we have a quadratic equation in r. Let's solve it using the quadratic formula:

r = (-b ± √(b^2 - 4ac)) / (2a),

where a = 1, b = -4, and c = 4.

Substituting the values into the formula, we get:

r = (-(-4) ± √((-4)^2 - 4(1)(4))) / (2(1)),

which simplifies to:

r = (4 ± √(16 - 16)) / 2,

r = (4 ± √0) / 2,

r = (4 ± 0) / 2.

This gives us two identical roots:

r = 2.

Therefore, the general solution to the differential equation is:

у = C1e^(2t) + C2te^(2t),

where C1 and C2 are arbitrary constants.

This is the solution to the given differential equation у'' - 4у' + 4у = 0.