7. Antiderivatives & Indefinite Integrals

Solve the initial value problems in Exercises 53–56 for y as a function of x.

(x² + 1)² (dy/dx) = √(x² + 1), where y(0) = 1

In Exercises 129–132 solve the initial value problem.

129. dy/dx = e^(-x-y-2), y(0) = -2

In Exercises 129–132 solve the initial value problem.

131. x dy - (y + √y)dx = 0, y(1) = 1

Solve the initial value problems in Exercises 67–70 for x as a function of t.

(t + 1) (dx/dt) = x² + 1 (for t > -1), x(0) = 0

Solve the initial value problems in Exercises 71–90.

y⁽⁴⁾ = −sin t + cos t;

y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0