Question

(e^x-y-e^-y)dx+(e^x+y+e^x)dy=0

Answer 1

$(e^{x-y}-e^{-y})dx+(e^{x+y}+e^{x})dy=0\\e^{-y}(e^{x}-1)dx + e^{x}(e^{y}+1)dy =0\\e^{-y}(e^{x}-1)dx=-e^{x}(e^{y}+1)dy\\e^{-x}(e^{x}-1)dx=-e^{y}(e^{y}+1)dy$

Далее интегрируем

$\int (1-e^{-x}) \, dx = -\int (e^{2y}+e^y) \, dy\\x+e^{-x}+C = -(\frac{e^{2y}}{2}+e^y)$

Обозначим: $f=e^y\\A = x+e^{-x}+C$

Тогда имеем квадратное уравнение с переменной f:

$A = -(\frac{f^2}{2}+f)\\f^2+2f+2A=0\\D=4-4*2A=4(1-2A)\\f_{1,2} = \frac{-2 \pm \sqrt{4(1-2A)}}{2} = -1 \pm \sqrt{1-2A} = -1 \pm \sqrt{1-2(x+e^{-x}+C)} = e^y$

Отрицательный корень отбрасываем, т.к. f=exp(y)>0.

Отсюда находим решение:

$y=\ln (-1+\sqrt{1-2(x+e^{-x}+C)})$

Answer 2

To solve the given first-order differential equation:

(e^x - y - e^-y)dx + (e^x + y + e^x)dy = 0

We can check if it's an exact differential equation by calculating its partial derivatives with respect to x and y.

∂(e^x - y - e^-y)/∂y = -1 - e^-y ∂(e^x + y + e^x)/∂x = e^x + e^x = 2e^x

The equation is exact if ∂M/∂y = ∂N/∂x, where M and N are the coefficients of dx and dy, respectively. In this case, it's not exact because ∂M/∂y ≠ ∂N/∂x.

To make it exact, we can find an integrating factor (μ(x, y)) such that:

μ(x, y) * [∂M/∂y - ∂N/∂x] = μ(x, y) * (-1 - e^-y - 2e^x) = 0

We can find a suitable μ(x, y) by equating the coefficient of μ(x, y) to zero:

-1 - e^-y - 2e^x = 0

Now, let's solve for μ(x, y):

1 + e^-y = 2e^x

e^-y = 2e^x - 1

Now, take the natural logarithm on both sides:

-y = ln(2e^x - 1)

y = -ln(2e^x - 1)

Now, we have μ(x, y) = e^(-ln(2e^x - 1)) = 1 / (2e^x - 1)

Multiply the entire equation by this integrating factor to make it exact:

1 / (2e^x - 1) * [(e^x - y - e^-y)dx + (e^x + y + e^x)dy] = 0

Now, we can check if it's exact:

∂/∂y[1 / (2e^x - 1) * (e^x - y - e^-y)] = ∂/∂x[1 / (2e^x - 1) * (e^x + y + e^x)]

You can compute these partial derivatives, and you'll find that they are equal, making the equation exact.

Now that it's exact, we can find the solution by integrating. To do this, we'll first find the potential function U(x, y):

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

Now, integrate the first term with respect to x:

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

U(x, y) = ∫(1 / (2e^x - 1) * (e^x - y - e^-y))dx + g(y)

Integrate the left-hand side with respect to x:

U(x, y) = ∫(1 / (2e^x - 1) * (e^x - y - e^-y))dx + g(y)

U(x, y) = ∫(1 / (2e^x - 1) * (e^x - y - e^-y))dx + g(y)

Now, we'll find the potential function U(x, y) by integrating the right-hand side with respect to x:

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

Now, integrate the left-hand side with respect to x:

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

Now, integrate the right-hand side with respect to x:

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

U(x, y) = ∫[1 / (2e^x - 1) * (e^x - y - e^-y)]dx + g(y)

The exact solution for the differential equation is given by:

U(x, y) = Constant

So, the solution to the differential equation is:

1 / (2e^x - 1) * (e^x - y - e^-y) + g(y) = Constant

Where g(y) is an arbitrary function of y.