What is the general solution to the differential equation given by xdydx = 4y + x^3 + x?

To find the general solution to the differential equation xdydx = 4y + x³ + x, we can begin by rearranging the terms and separating the variables.

First, we rewrite the equation:

xdydx - 4y = x3 + x.

This can be reorganized as:

xdydx = 4y + x3 + x.

Next, we divide both sides by x (assuming x ≠ 0) to make it easier to work with:

dydx = (4y/x) + x2 + 1.

This form shows that it’s a first-order linear differential equation. We can express it as:

dydx - (4/x)y = x2 + 1.

Now, we can identify the integrating factor μ(x), which is given by:

μ(x) = exp(∫ -4/x \, dx) = exp(-4 ln|x|) = |x|^{-4}.

Next, we multiply through by the integrating factor:

|x|^{-4}dydx - 4|x|^{-5}y = |x|^{-4}(x2 + 1).

After simplifying, we find:

(|x|^{-4}y)' = |x|^{-4}(x2 + 1).

Now, we can integrate both sides:

∫ (|x|^{-4}y)' dx = ∫ |x|^{-4}(x2 + 1) dx.

Calculating the right side, we can break it into two separate integrals:

∫ |x|^{-4}(x2) dx + ∫ |x|^{-4} dx.

Using the power rule, we have:

y|-3/x^3 + C - (1/3) |x|^{-3} + C_1

Where C is the constant of integration. Thus:

y = -rac{3C}{x^3} - rac{1}{3x^2} + C_1x4.

Therefore, the final general solution to the differential equation is:

y = -rac{3}{x^3} + C_1x4 + C

where C is an arbitrary constant. This solution is applicable for x > 0 and x < 0.