Another second-order equation Consider the differential equation y″ ( t ) + k 2 y ( t ) = 0, where k is a positive real number. a. Verify by substitution that when k = 1, a solution of the equation is y ( t ) = C 1 sin t + C 2 cos t. You may assume that this function is the general solution. b. Verify by substitution that when k = 2, the general solution of the equation is y ( t ) = C 1 sin 2 t + C 2 cos 2 t. c. Give the general solution of the equation for arbitrary k > 0 and verify your conjecture.
Another second-order equation Consider the differential equation y″ ( t ) + k 2 y ( t ) = 0, where k is a positive real number. a. Verify by substitution that when k = 1, a solution of the equation is y ( t ) = C 1 sin t + C 2 cos t. You may assume that this function is the general solution. b. Verify by substitution that when k = 2, the general solution of the equation is y ( t ) = C 1 sin 2 t + C 2 cos 2 t. c. Give the general solution of the equation for arbitrary k > 0 and verify your conjecture.
Solution Summary: The author explains that the given function y(t)=C_1mathrm
Another second-order equation Consider the differential equation y″(t) + k2y(t) = 0, where k is a positive real number.
a. Verify by substitution that when k = 1, a solution of the equation is y(t) = C1 sin t + C2 cos t. You may assume that this function is the general solution.
b. Verify by substitution that when k = 2, the general solution of the equation is y(t) = C1 sin 2t + C2 cos 2t.
c. Give the general solution of the equation for arbitrary k > 0 and verify your conjecture.
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
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