In Exercises 5–8, show that each function is a solution of the given initial value problem.
5. Differential Equation: 2y + y' = 4x + 2
Initial condition: y(-1) = e² - 2
Solution candidate: y = e^(-2x) + 2x

In Exercises 5–8, show that each function is a solution of the given initial value problem.
5. Differential Equation: 2y + y' = 4x + 2
Initial condition: y(-1) = e² - 2
Solution candidate: y = e^(-2x) + 2x
Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
73. lim (x → ∞) (2^x - 3^x) / (3^x + 4^x)
In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = e^(5-7x)
In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
29. y = (1 - t)coth⁻¹(√t)
Solve the initial value problems in Exercises 87 and 88.
88. d²y/dx² = sec²x, y(0)=0 and y'(0)=1
Indeterminate Powers and Products
Find the limits in Exercises 53–68.
60. lim (x → 0) (e^x + x)^(1/x)