Question

Y'/y=x в квадрате +x​

Answer 1

Ответ:

Пошаговое объяснение:

это уравнение с разделяющимися переменными

y'/y = x² +x

проинтегрируем обе части

$\displaystyle \int{\frac{dy/dx}{y} } \, dx =\int {(x^2+x)} \, dx$

ln(y) = x³/3 +x²/2 +C

$\displaystyle y=e^{x^3/3+x^2/2+C}$

переопределим константу

$\displaystyle y=C_1e^{x^3/3+x^2/2}$

Answer 2

It seems like you've written a differential equation. The notation $\frac{dy}{dx}$ represents the derivative of $y$ with respect to $x$. The equation you've provided is:

$\frac{dy}{dx} = x^2 + x$

This is a first-order ordinary differential equation. To solve it, you can use various methods, such as separation of variables, integrating factors, or direct integration.

Let's solve it using separation of variables:

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

Now, integrate both sides with respect to $x$:

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

The left side can be integrated as $\ln|y|$, and the right side can be integrated term by term.

$\ln|y| = \frac{1}{3}x^3 + \frac{1}{2}x^2 + C$

Here, $C$ is the constant of integration.

To solve for $y$, you can exponentiate both sides:

$|y| = e^{\frac{1}{3}x^3 + \frac{1}{2}x^2 + C}$

So, the general solution is:

$y = \pm e^{\frac{1}{3}x^3 + \frac{1}{2}x^2 + C}$

This is the general form of the solution to the given differential equation.