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
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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
Indeterminate Powers and Products
Find the limits in Exercises 53–68.
60. lim (x → 0) (e^x + x)^(1/x)
In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.
120. y = x^(sin x)
Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.
sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞
cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1
tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1
sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1
csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1
coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1
Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.
65. sech⁻¹(3/5)
In Exercises 139–142, find the length of each curve.
141. y = ln(cos(x)) from x = 0 to x = π/4.
In Exercises 27–32, find dy/dx.
ln y = e^y sinx