dy/dx = y^2/x^3
dy/y^2 = dx/x^3
-1/y = -1/(2x^2)+c
y = 2x^2/(1-cx^2)
what's with all the verbal noise? You want to discuss
y' = 2x√(1-y^2)? Just say so
dy/√(1-y^2) = 2x dx
arcsin(y) = x^2 + c
y = sin(c+x^2)
1.Solve the differential equation dy/dx= y^2/x^3 for y=f(x) with the condition y(1) = 1.
2.Solve the differential equation y prime equals the product of 2 times x and the square root of the quantity 1 minus y squared.
Explain why the initial value problem y prime equals the product of 2 times x and the square root of the quantity 1 minus y squared with y(0) = 3 does not have a solution.
3.The table below gives selected values for the function f(x). Use a trapezoidal estimation, with 6 trapezoids to approximate the value of the integral from 1 to 2 of f of x, dx. Give 3 decimal places for your answer.
X 1, 1.1, 1.2, 1.5, 1.7, 1.9, 2.0
f(x)1, 2, 4, 6, 7, 9, 10
2 answers
dy/dx=(y^2)(x^3)
1 answer