Question #351041

2.2. Solve the ivp sin(x) dx + y dy = 0, where y(0)

Expert's answer

"y dy = -\\sin(x) dx"

"y^2\/2=\\cos(x)+C"

"2=\\cos(0)+C=1+C"

"C=1"

Finally

"y^2=2\\cos(x)+2""y=\\pm\\sqrt{2\\cos(x)+2}"

The initial condition requres negative value of "y" for specified "x", thus

"y=-\\sqrt{2\\cos(x)+2}"

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