The general solution of the ode dydx+1xy=2x2𝑑𝑦𝑑𝑥+1𝑥𝑦=2𝑥2 isa.yx=x42+c𝑦𝑥=𝑥42+𝑐b.yx=x42𝑦𝑥=𝑥42c.yx=x33+c𝑦𝑥=𝑥33+𝑐d.yx=x44+c

The question seems to be a bit jumbled, but it appears to be asking for the general solution of a first order linear ordinary differential equation (ODE). The general form of such an equation is dy/dx + P(x)y = Q(x), where P(x) and Q(x) are some functions of x.

From the given equation, we can see that P(x) = 1/x and Q(x) = 2x^2.

The integrating factor e^(∫P(x)dx) = e^(∫1/x dx) = e^(ln|x|) = |x|.

Multiplying every term of the given differential equation by the integrating factor gives:

x dy/dx + y = 2x^3.

This is now in the form (d/dx)[x*y] = 2x^3, which can be integrated to give:

x*y = ∫2x^3 dx = 0.5x^4 + C.

So, the solution of the differential equation is y = 0.5x^3 + C/x, which is not exactly in the form of any of the provided options. There might be a mistake in the question or the provided options.