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## Main Question or Discussion Point

The solution of the initial value problem, dy/dx = y-3. y(0) = 4, what does y=?

dy/y-3 = dx

1/y-3 dy = dx

ln(y-3) = ? + C

I have no idea how to do this one, can anybody please help me out? I'm having a hard time seeing how to separate the y and the -3 or if they are separable at all.

dy/y-3 = dx

1/y-3 dy = dx

ln(y-3) = ? + C

I have no idea how to do this one, can anybody please help me out? I'm having a hard time seeing how to separate the y and the -3 or if they are separable at all.