Solve the initial value problem yy' + x = sqrt(x^2 + y^2) with y(4) = -sqrt(33).
a. To solve this, we should use the substitution
u = x^2 + y^2
u' = 2(x+yy')
Enter derivatives using prime notation (e.g., you would enter y' for dy/dx).
b. After the substitution from the previous part, we obtain the following linear differential equation in x, u, u'.
u' = 2(u)^(1/2)
c. The solution to the original initial value problem is described by the following equation in x, y.
y =sqrt((x+3)^2-x^2)