show that y= Ae^kx+C is a solution of y'=k(y-C)

Thank you

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- May 12th 2009, 01:16 AMslaypullingcatexponential differentiation
show that y= Ae^kx+C is a solution of y'=k(y-C)

Thank you - May 12th 2009, 01:52 AMSpec
$\displaystyle \frac{d}{dx}\left(Ae^{kx}+C \right)=k\cdot Ae^{kx}$