Question

Y''-3y'-4y=0 ================

Answer 1

$y''-3y'-4y=0$

Составляем характеристическое уравнение:

$\lambda^2-3\lambda-4=0$

$D=(-3)^2-4\cdot1\cdot(-4)=25$

$\lambda_1=\dfrac{3+\sqrt{25} }{2} =4$

$\lambda_2=\dfrac{3-\sqrt{25} }{2} =-1$

Так как корни характеристического уравнения - различные действительные числа, то решение уравнения запишется в виде:

$y=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$

Решение уравнения:

$\boxed{y=C_1e^{4x}+C_2e^{-x}}$

Answer 2

Ответ:

$y''-3y'-4y=0$

Характеристическое уравнение:   $k^2-3k-4=0\ \ \ \Rightarrow \ \ \ k_1=-1\ ,\ k_2=4\ \ (teorema\ Vieta)$

Получили два различных действительных корня характеристического уравнения, поэтому общее решение записываем в виде:

$y=C_1e^{-x}+C_2e^{4x}$

Answer 3

The equation you provided is a second-order linear homogeneous differential equation:

Y'' - 3Y' - 4Y = 0

To solve this differential equation, we can assume a solution of the form Y = e^(rt), where r is a constant to be determined.

Let's substitute this assumed solution into the differential equation:

Y'' - 3Y' - 4Y = 0 (e^(rt))'' - 3(e^(rt))' - 4(e^(rt)) = 0

Differentiating e^(rt) with respect to t gives:

r^2e^(rt) - 3re^(rt) - 4e^(rt) = 0

Now we can factor out e^(rt):

e^(rt)(r^2 - 3r - 4) = 0

Since e^(rt) is never zero, the term in the parentheses must equal zero:

r^2 - 3r - 4 = 0

Now we can solve this quadratic equation for r. Factoring the equation gives:

(r - 4)(r + 1) = 0

Setting each factor equal to zero:

r - 4 = 0 or r + 1 = 0

r = 4 or r = -1

We have two possible values for r. This means we have two linearly independent solutions.

For r = 4, the solution is:

Y1(t) = e^(4t)

For r = -1, the solution is:

Y2(t) = e^(-t)

Therefore, the general solution to the given differential equation is:

Y(t) = c1e^(4t) + c2e^(-t)

where c1 and c2 are arbitrary constants that can be determined based on initial conditions or additional information.