Find the general solution to the differential equation

ani9890
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Find the general solution to the differential equation
y'+(12x^11)y=x^12

Use the variable I= the integral of e^(x^12)dx where it occurs in your answer.

According to some people, it doesn't have an elementary solution, look at:
http://www.wolframalpha.com/input/?i=y%27+%2B+12*x^11*y+%3D+x^12

there's a incomplete gamma function ?

can someone please show me how to solve this problem, thank you so much!
 
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Hi ani9890! Welcome to PF :smile:

Wolfram Alpha has a strange way of solving differential equations. But there is an elementary solution, as it is a simple linear differential equation. How can you solve such equations?

Hint : Start by analyzing what the integrating factor is...
 
so I end up with this,

d/dx[ye^(x^12)] = x^12*e^(x^12)

and I'm supposed to integrate both sides, but the right side can't be integrated to give a elementary solution. So what should I do?

I was thinking y(x)=e^(-x^12)(x^12)I
where I stands for the integral of e^(x^12) , but I'm sure this is wrong lol
 
ani9890 said:
so I end up with this,

d/dx[ye^(x^12)] = x^12*e^(x^12)

and I'm supposed to integrate both sides, but the right side can't be integrated to give a elementary solution. So what should I do?

I was thinking y(x)=e^(-x^12)(x^12)I
where I stands for the integral of e^(x^12) , but I'm sure this is wrong lol

No:

\displaystyle \frac{d}{dx}\left(ye^{\displaystyle x^{12}}\right)=y'\ e^{\displaystyle x^{12}}+y\,e^{\displaystyle x^{12}}\left(12x^{11}\right)

So, what if you multiply your Dif.Eq. by e12x ?
 
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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