Find the general solution of the second-order, linear, homogeneous differential equation 4y′′− 20y′ + 25y = 0.

I have got up to the stage where I have found landa to be =5/2 but do not know what to do after that. Please help.

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- Feb 21st 2009, 12:51 PMronaldo_07[SOLVED] Differential Equation
Find the general solution of the second-order, linear, homogeneous differential equation 4y′′− 20y′ + 25y = 0.

I have got up to the stage where I have found landa to be =5/2 but do not know what to do after that. Please help. - Feb 21st 2009, 01:02 PMJester
If $\displaystyle y = e^{mx}$ then $\displaystyle 4m^2-20m+25 = 0$ and as you said $\displaystyle m = \frac{5}{2}, \frac{5}{2}$. Sicne the root is repeated the the two solutions are

$\displaystyle y_1 = e^{\frac{5x}{2}},\;\;\;y_2 = x\,e^{\frac{5x}{2}} $

and the general solution solution

$\displaystyle y = c_1 e^{\frac{5x}{2}} + c_2 x\,e^{\frac{5x}{2}} $ - Feb 21st 2009, 01:05 PMronaldo_07
I got that already, is that the final answer? Do you not need to find C1 and C2? and where did the x come from in y2?

- Feb 21st 2009, 02:26 PMJester
- Feb 28th 2009, 04:07 PMronaldo_07
- Feb 28th 2009, 05:06 PMHallsofIvy
You have a "general solution" with two unknown constants and you have two additional conditions. Put your general solution into those conditions and you will have two equations to solve for the two constants.