Initial Value Problem
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Which of the following is a solution of the given initial-value problem.
y' + tan(x)
y = 4 cos^2(x)
y(0) = -4.
on the interval -pi/2 < x < pi/2
A) y = 4 sin(x) - 4 cos(x)
B) y = 4 cos(x) - 4 sin(x)cos(x)
C) y = 4 sin(x)cos(x) + 4 cos(x)
D) y = 4 sin(x)cos(x) - 4 cos(x)
E) y = 4 sin(x) + 4 cos(x)
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Solution Summary
Initial value function problems are solved for different intervals.
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Answer: D
First, I check the initial value.
A. y(0)=-4 B. y(0)=4 C. y(0)=4 D. y(0)=-4 E. y(0)=4
The condition says y(0)=-4, so only A and D are possible answers.
Second, I check the equation.
A. y = 4sin(x) - 4cos(x), ...
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