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I’m having some difficulty understanding the solution to the following differential equation problem.

Find a general solution to the given differential equation

$4y’’ – 4y’ + y = 0$

The steps I’ve taken in solving this problem was to first find the auxiliary equation and then factor to find the roots. I listed the steps below:

$4r^2 – 4r + 1$

$(2r – 1) \cdot (2r-1)$

$\therefore r = \frac{1}{2} \text{is the root}$

Given this information, I supposed that the general solution to the differential equation would be as follows:

$y(t) = c_{1} \cdot e^{\frac{1}{2} t}$

But when I look at the back of my textbook, the correct answer is supposed to be

$y(t) = c_{1} \cdot e^{\frac{1}{2} t} + c_{2} \cdot te^{\frac{1}{2} t}$

Now I know that understanding the correct solution has something to do with linear independence, but I’m having a hard time getting a deep understanding of what’s going on. Any help would be appreciated in understanding the solution.

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up vote 2 down vote accepted

The story behind what is going on here, is exactly the same we always see when search a Basis for a vector space over $V$ a field $K$. There; we look for a set of linear independent vectors which can generate the whole space. In the space of all solutions for an OE, we do the same as well. For any Homogeneous Linear OE with constant coefficients, there is a routine way to find out the solutions. And you did it right for this one. When the solution of axillary equation is one and this solution frequents two times, one solution is as you noted $y_1(t)=\exp(0.5t)$ but we don't have another solution. So, we should find another solution which is independent to the first one and the number of whole solutions is equal to the order of OE which is two here. It means that, we need one solution that the set $$\{\exp(0.5t),y_2(t)\}$$ is a fundamental set of solutions. For doing this you can use the method of Reduction of Order to find $$y_2(t)=t\exp(0.5t)$$

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I like your thought process - it helps to show the "why", not just "what" the answer is! +1 – amWhy Feb 4 '13 at 1:03

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