Question

(e^2x =1) dy =ye^2x dx=0

Answer 1

Ответ:

Пошаговое объяснение:

Уравнение с разделяющимися переменными.

Разделяем переменные

y·e2xdx=(1+e2x)dy

Делим обе части на у·(1+e2x)

e2xdx/(1+e2x)=dy/y

Интегрируем

∫ dy/y= ∫ e2xdx/(1+e2x)

Применяем подведение под дифференциал в интеграле справа

∫ dy/y= 1/2∫ d(1+e2x/(1+e2x)

ln|y|=(1/2)ln(1+e2x)dx+lnC

y=C·√1+e2x – о т в е т.

Answer 2

It looks like you have provided a system of differential equations:

1. $e^{2x} \frac{dy}{dx} = y e^{2x}$
2. $dy = 0$

Let's work through this system of equations:

Starting with the second equation, $dy = 0$, this simply states that the derivative of $y$ with respect to $x$ is zero. This means that $y$ is a constant with respect to $x$, and its derivative is zero.

Moving on to the first equation, $e^{2x} \frac{dy}{dx} = y e^{2x}$, we can divide both sides by $e^{2x}$ since $e^{2x}$ is never zero:

$\frac{1}{e^{2x}} \frac{dy}{dx} = y$

Now, notice that the left-hand side can be simplified further:

$\frac{d}{dx} \left(\frac{1}{e^{2x}} y\right) = y$

Integrating both sides with respect to $x$ gives:

$\int \frac{d}{dx} \left(\frac{1}{e^{2x}} y\right) dx = \int y dx$

$\frac{1}{e^{2x}} y = \int y dx + C$

Here, $C$ is the constant of integration. Solving for $y$:

$y = e^{2x} \left(\int y dx + C\right)$

Since we know that $\frac{dy}{dx} = 0$, this implies that $y$ is a constant, let's say $y = K$:

$K = e^{2x} \left(\int K dx + C\right)$

Solving for $C$:

$C = K - \int K dx$

So, the general solution for the given system of differential equations is:

$y = e^{2x} \left(K - \int K dx\right)$

Here, $K$ and $C$ are constants. The equation represents a family of solutions, parameterized by $K$, that satisfy the given differential equations.