S sunny1324 New member Joined Jul 12, 2008 Messages 26 Jul 26, 2008 #1 Find the general solution of the following equation y'' +10y'+25y=14e^(-5x) r^2+10r+25=0 r1=-5, r2=-5 Yh = Ae^(-5X) + Bxe^(-5x) Yp= ax^2 * e^(-5x) how do I find what the derivative of Yp is?
Find the general solution of the following equation y'' +10y'+25y=14e^(-5x) r^2+10r+25=0 r1=-5, r2=-5 Yh = Ae^(-5X) + Bxe^(-5x) Yp= ax^2 * e^(-5x) how do I find what the derivative of Yp is?
D Deleted member 4993 Guest Jul 26, 2008 #2 Re: Undetermined Coefficients sunny1324 said: Find the general solution of the following equation y'' +10y'+25y=14e^(-5x) r^2+10r+25=0 r1=-5, r2=-5 Yh = Ae^(-5X) + Bxe^(-5x) Yp= ax^2 * e^(-5x) how do I find what the derivative of Yp is? <<< By applying product rule of differentiation. Click to expand...
Re: Undetermined Coefficients sunny1324 said: Find the general solution of the following equation y'' +10y'+25y=14e^(-5x) r^2+10r+25=0 r1=-5, r2=-5 Yh = Ae^(-5X) + Bxe^(-5x) Yp= ax^2 * e^(-5x) how do I find what the derivative of Yp is? <<< By applying product rule of differentiation. Click to expand...
S sunny1324 New member Joined Jul 12, 2008 Messages 26 Jul 27, 2008 #3 Re: Undetermined Coefficients oh right. i forgot about that, thanks
