Solve the following ODEs:(i) xyy′ = y2 + 2x2 with initial conditions y(1) = 2

The given ordinary differential equation (ODE) is xyy′ = y^2 + 2x^2.

Step 1: Rewrite the ODE in the standard form. The standard form of a first order ODE is y′ = f(x, y). So, we rewrite the given ODE as y′ = (y^2 + 2x^2) / xy.

Step 2: Check if the ODE is separable. A separable ODE can be written in the form of g(y)y′ = h(x). In this case, the ODE is not separable.

Step 3: Check if the ODE is exact. An exact ODE can be written in the form M(x, y) + N(x, y)y′ = 0, where M_y = N_x. In this case, the ODE is not exact.

Step 4: Check if the ODE is linear. A linear ODE can be written in the form y′ + p(x)y = g(x). In this case, the ODE is not linear.

Step 5: Check if the ODE is homogeneous. A homogeneous ODE can be written in the form y′ = f(y/x). In this case, the ODE is homogeneous.

Step 6: Solve the homogeneous ODE. The general solution of a homogeneous ODE is y = vx, where v is a function of x. Substituting y = vx into the ODE, we get x^2v^2 + 2x^2 = vx^2v′. Simplifying, we get v = v′ + 2/v. This is a separable ODE, which can be solved by integrating both sides with respect to x.

Step 7: Apply the initial condition. The initial condition is y(1) = 2. Substituting x = 1 and y = 2 into the general solution, we can solve for the constant of integration.

Step 8: Write the final solution. The final solution is the general solution with the constant of integration replaced by the value obtained in step 7.