why is the derivative of \[(Ax+B)e^x\] \[(Ax+A+B)e^x\]? I understand the Ax+A part, but why didn't the B disappear? btw it's an ODE problem. \[y''-4y=xe^x+cos2x\]

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why is the derivative of \[(Ax+B)e^x\] \[(Ax+A+B)e^x\]? I understand the Ax+A part, but why didn't the B disappear? btw it's an ODE problem. \[y''-4y=xe^x+cos2x\]

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\[y=(Ax+B)e^x\]\[y'=(Ax+B)e^x+Ae^x=(Ax+A+B)e^x\]
how odd, wolfram alpha won't show me the steps unless I sign up with them.
why doesn't the B disappear? Isn't it a constant?

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Other answers:

  • phi
derivative of B * e^x is B* d/dx (e^x) = B* e^x
By the product rule the B disappears from one Ax+B term, and in the other term you differentiate e^x \[u=Ax+B\]\[u'=A\]\[v=e^x\]\[v'=e^x\]\[uv'+vu'=(Ax+B)e^x+Ae^x\]factor out \(e^x\) and you get\[(Ax+A+B)e^x\]
awesome thanks!!!!

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