Question

X'+2xy=-xe^-x^2 sin x, y=(0)

Answer 1

Ответ:

в условии опечатки и начальных условий нет. так что частное решение сами найдёте

Answer 2

It looks like you've provided a first-order linear ordinary differential equation (ODE) and want to find the solution for the given initial condition. The equation is:

$x' + 2xy = -xe^{-x^2} \sin x$

Where $y(0) = 0$. To solve this ODE, we'll use an integrating factor method. The general form of a first-order linear ODE is $y' + P(x)y = Q(x)$.

1. Identify $P(x)$ and $Q(x)$: In your equation, $P(x) = 2x$ and $Q(x) = -xe^{-x^2} \sin x$.

2. Calculate the integrating factor: The integrating factor $\mu(x)$ is given by $\mu(x) = e^{\int P(x) \, dx}$. In this case, $\int P(x) \, dx = \int 2x \, dx = x^2$, so $\mu(x) = e^{x^2}$.

3. Multiply the entire equation by $\mu(x)$: Multiplying the equation by the integrating factor $e^{x^2}$ gives: $e^{x^2}x' + 2xe^{x^2}y = -x \sin x e^{x^2 - x^2}$ Simplifying: $(e^{x^2}x)' = -x \sin x$

4. Integrate both sides: $\int (e^{x^2}x)' \, dx = -\int x \sin x \, dx$ $e^{x^2}x = \int x \sin x \, dx + C$

5. Solve for $x$: At this point, you'll need to evaluate the integral on the right-hand side, which can be done using integration by parts. After that, solve for $x$.

6. Solve for $y$: Once you've found the expression for $x$, you can substitute it back into the equation $y = \int \frac{-x \sin x}{e^{x^2}} \, dx$ and evaluate the integral. This will give you the expression for $y$ in terms of $x$.

7. Apply the initial condition: Finally, substitute $x = 0$ into the expression for $y$ to find the value of the constant of integration, ensuring that $y(0) = 0$.

Remember that the integral involving $\sin x$ and $e^{x^2}$ might not have a simple closed-form solution, and you might need to use numerical methods to approximate the solution.