Third order Euler Differential Equation

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Homework Statement


x^3y'''-3x^2y''+7xy'-8y=x+e^2x

Edit : Got it thanks
 
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Have you thought about trying the solution y=x^{n}? This might lead to an easier cubic to solve.
 
hunt_mat said:
Have you thought about trying the solution y=x^{n}? This might lead to an easier cubic to solve.

Not sure how that would work out, only told to use t=lnx. But using y=x^{n} won't cancel the coefficients to be constants.

Edit: I looked up your response on Wikipedia and found something similar on it, never learned to do it this way tho. I'll give it a shot.
 
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They will add up to give x^{n} though. for example if y=x^n, then x^{3}y'''(x)=n(n-1)(n-2)x^n, likewise for x^2 and so on. I calculate that the cubic you arrive at the cubic:
<br /> n^{3}-6n^{2}+12n-8<br />
One of the solutions look to be n=2.

Mat
 
hunt_mat said:
They will add up to give x^{n} though. for example if y=x^n, then x^{3}y'''(x)=n(n-1)(n-2)x^n, likewise for x^2 and so on. I calculate that the cubic you arrive at the cubic:
<br /> n^{3}-6n^{2}+12n-8<br />
One of the solutions look to be n=2.

Mat

Got it thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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