$\displaystyle d^2y/dx^2-7 dy/dx +6y=36x $

given that when y=0 $\displaystyle ,dy/dx=4$ and x=0

i would appreciate some help with this ones is all new to me. what do i do with the variables given.

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- Apr 8th 2010, 10:22 AMsigma1Non-homogeneous 2nd order DE with constant coefficients.
$\displaystyle d^2y/dx^2-7 dy/dx +6y=36x $

given that when y=0 $\displaystyle ,dy/dx=4$ and x=0

i would appreciate some help with this ones is all new to me. what do i do with the variables given. - Apr 8th 2010, 10:52 AMRandom Variable
It's a nonhomogeneous linear equation. So you'll need to find a particular solution of the nonhomogeneous equation and the general solution of the related homogeneous equation

The homogeneous equation is $\displaystyle \frac{d^{2}y}{dx^{2}} - 7 \frac{dy}{dx} + 6y = 0 $

The characteristic equation is $\displaystyle r^{2}-7x+6 =0 $, which has roots of r=6 and r =1.

So the general solution of the homogeneous equation is $\displaystyle y_{h}(x) = C_{1}e^{6x}+C_{2}e^{x} $

Now we need any particular solution of $\displaystyle \frac{d^{2}y}{dx^{2}} - 7 \frac{dy}{dx} + 6y = 36 $

$\displaystyle y_{p}(x)=6 $ will do

so the general solution on the nonhomongeneous equation is $\displaystyle y(x) = y_{h}(x)+y_{p}(x) = C_{1}e^{6x}+C_{2}e^{x} + 6 $

Now we need to find the two coefficients by using the initial conditions.

$\displaystyle y(0)=0=C_{1}+C_{2} + 6$

$\displaystyle \frac{dy}{dx} = 6C_{1}e^{6x}+C_{2}e^{x} $

$\displaystyle \frac{dy}{dx} (0) = 4= 6C_{1}+C_{2} $

Solving the two equations simultaneously, $\displaystyle C_{1}= 2$ and $\displaystyle C_{2}= -8 $

so our final solution is $\displaystyle y(x)= 2e^{6x}-8e^{x}+6 $ - Apr 8th 2010, 12:15 PMsigma1
thanks alot i was not familar with the solution for the general solution on the nonhomongeneous equation. thanks alot i really appreciate it.