# Second order non linear inhomogeneous differential equation.. :(

#### Khonics89

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
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

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

Why assume x=0?