Question

4d²y/dx²+4dy/dx+y=0 help me please​

Answer 1

Ответ:       y = ( C₁ + C₂x )e^( - 0,5x) .

Пошаговое объяснение:

4d²y/dx²+4dy/dx+y=0 ;  перепишемо так :

4у '' + 4y ' + y = 0 ;  його характеристичне рівняння

4k² + 4k + 1 = 0 ;

(  2k + 1 )² = 0 ;      2k + 1 = 0 ;   2k = - 1 ;     k = - 0,5 ;

k₁ = k₂ = - 0,5 ;

y = ( C₁ + C₂x )e^( - 0,5x) .

Answer 2

Ответ:

пререведи потом я дам ответ

Answer 3

Certainly! The given differential equation is a second-order linear homogeneous differential equation with constant coefficients. It can be written in the standard form:

$4 \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} + y = 0$

To solve this differential equation, you can assume a solution of the form $y = e^{rx}$, where $r$ is a constant. Then, find the derivatives and substitute them into the differential equation:

$y = e^{rx}$ $\frac{dy}{dx} = re^{rx}$ $\frac{d^2y}{dx^2} = r^2e^{rx}$

Substitute these into the differential equation:

$4r^2e^{rx} + 4re^{rx} + e^{rx} = 0$

Now, factor out $e^{rx}$:

$e^{rx}(4r^2 + 4r + 1) = 0$

For a non-trivial solution (i.e., $e^{rx} \neq 0$), the expression in the parentheses must be zero:

$4r^2 + 4r + 1 = 0$

Now, solve this quadratic equation for $r$. You can use the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 4$, $b = 4$, and $c = 1$. Substitute these values into the formula and simplify to find the values of $r$. Once you have the values of $r$, the general solution to the differential equation is given by the linear combination of the solutions associated with each $r$:

$y(x) = c_1e^{r_1x} + c_2e^{r_2x}$

where $c_1$ and $c_2$ are arbitrary constants determined by the initial conditions if they are given.

Please note that the specific form of the solution depends on the nature of the roots of the quadratic equation. If the roots are real and distinct, the solution involves terms of the form $e^{r_1x}$ and $e^{r_2x}$. If the roots are real and equal, the solution involves terms like $e^{rx}$ and $xe^{rx}$. If the roots are complex, the solution involves trigonometric functions.