^{3}y'''. We then plug this into a polynomial and solve in order to find the coefficients to the general solution. Depending on the solutions of the roots of the polynomial we get different general solutions. Lets find the solution to the following equation.

Now lets take the information we need from this and plug it into the following equation which is the characteristic equation for an Euler Cauchy equation thats second order.

Where we get b and c knot from

So this means that if x

^{2}has some coefficient or didn't match this form, we'd have to adjust the equation to make the first term x^{2}. In this case though its fine, and we just have to find b(x) and c(x), which are
b(x) = -3

c(x) = 4

which means that our equation will be

Which after factoring becomes

So r is equal to 2 twice, which means its a repeated result, one of the possibilities. Now the general solution to the Euler Cauchy equation is x to the n where n is the solution, but in this case we have a repeated result. This means we need to add an additional factor of ln(x) into our solution giving us the following.

Which is our final result assuming there's no initial conditions.

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