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#1
10-19-2008, 07:29 PM
 ben489 Junior Member Join Date: Oct 2008 Posts: 1
Solving a 2nd order differential equation

Could anyone help me solve this differential equation:

my'' = -mg -ky'

with initial conditions:

y(0)=0
y'(0)=vo=usin(alpha)

If possible could you go through this step by step as i think i have made a minor error along the way...

Thanks
#2
10-21-2008, 02:52 PM
 HallsofIvy Senior Member Join Date: Oct 2006 Posts: 464

Quote:
 Originally Posted by ben489 Could anyone help me solve this differential equation: my'' = -mg -ky' with initial conditions: y(0)=0 y'(0)=vo=usin(alpha) If possible could you go through this step by step as i think i have made a minor error along the way... Thanks
Let x= y' and that equation becomes mx'= -mg-kx, a separable first order equation. dx/(mg+kx)= -1/m dt so, integrating, (1/k)ln(mg+ kx)= -t/m+ C or ln(mg+kx)= -kt/m+ kC. Taking the exponential of both sides, mg+ kx= C' e^(-kt/m) where C' is e^C.

Now x= y'= C"e^(-kt/m)- mg/k. Since y'(0)= C"- mg/k= v0, C"= (v0+ mg/k) and y'= (v0+mg/k)e^(-kt/m)- mg/k. Integrating again, y= (-m/k)(v0+ mg/k)e^(-kt/m)- mgt/k+ C= -(mv0/k+ m^2g/k^2)e^(-kt/m)- mgt/k+ C.

Since y(0)= -(mv0/k+ m^2g/k^2)+ C= 0, C= mv0/k+ m^2g/k^2.

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