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Initial Value Problem

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)

Solution Preview

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), ...

Solution Summary

IVPs are solved.

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