The image you uploaded contains five differential equations to solve. Let’s go step by step.

1. $x \frac{dy}{dx} = x^3 e^{3x} y^4 - 2x y^4$

This can be simplified by dividing both sides by $x$, which gives:

$\frac{dy}{dx} = x^2 e^{3x} y^4 - 2y^4$

At this point, we would need to separate the variables to solve this. You can proceed by factoring out $y^4$ on the right-hand side, then separate terms involving $y$ and $x$.

2. $xy' + 3y = x \sin(x)$

This is a first-order linear differential equation. We can solve it using the integrating factor method. The equation is in the form:

$y' + \frac{3}{x} y = \sin(x)$

To solve this, we compute the integrating factor $\mu(x)$:

$\mu(x) = e^{\int \frac{3}{x} dx} = x^3$

Multiplying through by $x^3$ and solving the resulting equation would yield the general solution.

3. $(2y^2 - 6xy - x)dx + (4xy - 3x^2 + e^y)dy = 0$

This is a first-order linear differential equation that we can attempt to solve by checking if it is exact, i.e., if $M(x,y)dx + N(x,y)dy = 0$ is exact (where $M$ and $N$ are the functions of $x$ and $y$).

First, check if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

4. $\left( x + x^2 y^2 + 4y \right) \frac{dy}{dx} = 0, \quad y(4) = 0$

This equation seems to be separable. You can first simplify it by dividing through by $\left( x + x^2 y^2 + 4y \right)$. This will allow you to solve it by separating the variables $x$ and $y$, and applying the initial condition $y(4) = 0$ to find the particular solution.

5. $(x^2 + y^2)dx = 2xy dy$

This is a separable equation. We can rearrange it as:

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

Then, separate the variables and integrate both sides to solve for $y(x)$.

Would you like a detailed solution for any of these steps? Feel free to ask for clarifications!