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2nd Order Non-Homogeneus please help driving crazy !
- Thread starter cihadkoc
- Start date
D
Deleted member 4993
Guest
2y2y" + 2y (y')2 = 1
Thats the equation. Which method should i use please help?
You are sure that the problem is NOT:
y2y" + 2y(y')2 = 1
D
Deleted member 4993
Guest
Yes I am pretty sure :/ But I am glad to listen the solution if its starting with y2
In that case, what are your thoughts?
Actually I dont have much idea. I thought you have a sloution for that and wanted to see that.
I did v=y' and conitnued
But nonhomogeneous is really hard in that case
2y2v dv/dy + 2yv2 - 1 = 0
then;
v ( 2y2 dv/dy + 2yv -1/v) =0
Now v= 0 or ( 2y2 dv/dy + 2yv -1/v)= 0
But I cant solve it.. :/ I am not sure if i am doing right also
What do you think ?
I did v=y' and conitnued
But nonhomogeneous is really hard in that case
2y2v dv/dy + 2yv2 - 1 = 0
then;
v ( 2y2 dv/dy + 2yv -1/v) =0
Now v= 0 or ( 2y2 dv/dy + 2yv -1/v)= 0
But I cant solve it.. :/ I am not sure if i am doing right also
What do you think ?
D
Deleted member 4993
Guest
I went to wolframalpha.com and fed the DE and got a horrendous answer.
D
Deleted member 4993
Guest
Yes I am pretty sure :/ But I am glad to listen the solution if its starting with y2
\(\displaystyle \dfrac{d}{dx}\left[y^2*y'\right] \ = \ ?\)
D
Deleted member 4993
Guest
So now you know how to solve the problem if it did not have '2' in the first term!
Its = 2y(y')2 + y2y''
Now its like the problem.. Bu unfortunately Its starting with 2.
Nothing new, just wondering if the problem has actually changed from the initially stated. If the problem is supposed to be
2 y (y')2 + y2 y'' = 1
then re-write as suggested by Subhotosh Khan
2 y (y')2 + y2 y'' = (y2 y')' = 1
and, as mentioned by cihadkoc, the final solution is
y(x) = \(\displaystyle (\frac{3}{2}\space x^2\space +\space 3\space c_1\space x\space +\space 3\space c_2)^{1/3}\)
Of course, you could just call the 3 c1 and 3 c2 new constants c3 and c4 and write
y(x) = \(\displaystyle (\frac{3}{2}\space x^2\space +\space c_3\space x\space +\space c_4)^{1/3}\)
2y2y" + 2y (y')2 = 1
Thats the equation. Which method should i use please help?
You know, I missed the perfect opportunity here. When someone says something is driving them crazy, the proper response is 'Wrong. It's only a short walk."