1. ## [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.

2. Originally Posted by ronaldo_07
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.
If $y = e^{mx}$ then $4m^2-20m+25 = 0$ and as you said $m = \frac{5}{2}, \frac{5}{2}$. Sicne the root is repeated the the two solutions are

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

and the general solution solution

$y = c_1 e^{\frac{5x}{2}} + c_2 x\,e^{\frac{5x}{2}}$

3. 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?

4. Originally Posted by ronaldo_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?
$c_1\;\;\text{and}\;\;c_2$ are found from some IC's. Have you not done repeated roots and reduction of order?

5. Originally Posted by ronaldo_07
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.

Sorry i missed out some info here is the rest:

Find the particular solution that satisfies the initial conditions y(0) = 2 and y′(0) = 1/2.

6. 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.