Recent content by song90

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    Solution for 2nd ODE Question 2d2ydx2 + 4 dydx+ 7y = e^−x cos x

    i was told by lecturer (maybe i misunderstanding what she said). If yp has the common with yc, yp=(equation)x. something like this... I have no idea about this lor. anyways thanks a lot for your help.
  2. S

    Solution for 2nd ODE Question 2d2ydx2 + 4 dydx+ 7y = e^−x cos x

    okay looked back on my workings i found that i had timed a "x" in the yp equation, yp=(e-x(ACcos x + BCsin x))x. i was thinking yp had common with yc as yc=e-x(Asin√(5/2)+Bcos√(5/2))
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    Solution for 2nd ODE Question 2d2ydx2 + 4 dydx+ 7y = e^−x cos x

    thanks for immediate reply. i was trying to solve the equations but i still couldn't get the answer for yp which is 1/3(e-xcos x). Wondering where the mistakes i had made
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    Solution for 2nd ODE Question 2d2ydx2 + 4 dydx+ 7y = e^−x cos x

    sorry, i am new to this forum... yp = Ce-x(Acos x + Bsin x). insert a constant at e-x since it is one of the members of undetermined coefficient? 2y'' + 4y' + 7y = e-xcos x <this is the original question and the answer for yp is 1/3(e-xcos x). Still i could not manage to get the ans for yp...
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    Solution for 2nd ODE Question 2d2ydx2 + 4 dydx+ 7y = e^−x cos x

    do e-x there put a constant? let say p=Ce[SUP]-x(Acos x + Bsin x)
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    Solution for 2nd ODE Question 2d2ydx2 + 4 dydx+ 7y = e^−x cos x

    2d2ydx2 + 4 dydx+ 7y = e^−x cos x i solved the equation for yc. but couldn't solve for yp as i dint know what kind of undetermined method to put for yp, particular integral.
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