Second order non linear inhomogeneous differential equation.. :(

Apr 2009
108
0
Solve

y''= x^3/y

I tried by using let u=y'

then I get

u'=x^3/y .. now something tells me this is not going to be an easy ride home.
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
Lets write the second order DE as...

\(\displaystyle y^{''} y = x^{3}\) (1)

... and suppose that the solution is analytic in \(\displaystyle x=0\) so that is...

\(\displaystyle y(x)= a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} + \dots \) (2a)

\(\displaystyle y^{'} (x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \dots + n a_{n} x^{n-1} + \dots\) (2b)

\(\displaystyle y^{''} (x) = 2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + \dots + n(n-1) a_{n} x^{n-2} + \dots\) (2c)

The \(\displaystyle a_{n}\) are found inserting the (2a) and (2c) in (1), once we know the 'initial conditions' \(\displaystyle y(0)=a_{0}\) and \(\displaystyle y^{'} (0) = a_{1}\) ...

\(\displaystyle 2 a_{0} a_{2} =0 \rightarrow a_{2} = 0\)

\(\displaystyle 6 a_{0} a_{3} + 2 a_{1} a_{2} = 0 \rightarrow a_{3}=0\)

\(\displaystyle 12 a_{0} a_{4} + 6 a_{1} a_{3} + 2a_{2}^{2}= 0 \rightarrow a_{4}=0\)

\(\displaystyle 20 a_{0} a_{5} + 12 a_{1} a_{4} + 8a_{2} a_{3}= 1 \rightarrow a_{5}= \frac{a_{0}}{20}\)

\(\displaystyle 30 a_{0} a_{6} + 20 a_{1} a_{5} + 12a_{2} a_{4} + 6 a_{3}^{2} = 0 \rightarrow a_{6}= - \frac{a_{1}}{30}\)

... and so one...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Why assume x=0?

Why not just start with u=y' ?
Because then you get a single equation with the two unknown functions, u and y.

And he did NOT "assume x= 0", he assumed the solution was analytic at x= 0.