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This is not a simple question.

           Solving    d²y/dx² = - x² y    --- (1)

This is in the form of Fowlers' second order non linear differential equation.

Solution is not known in the form of a simple expression in terms of functions. So we will do using Taylor series expansion form of infinite series for y (x).

y(x)=\Sigma_{n=0}^{\infty}\ \ a_n\ x^n      ---- (2)

As the second derivative y" is a multiple of x² of y itself, we can find two independent different solutions one series with even powers or x and one with odd powers of x.

When we differentiate y wrt x twice, the power of x in series reduces by 2 in eq(2).  But in equation (1), we have that y" has each term with power of x increased by two.  Hence, in the equation (2), we must have powers of x of successive terms with a difference of 4. So then:

y_1(x)=\Sigma_{n=0}^{\infty}\ a_{4n}\ x^{4n}\\ \. \ \ \ =a_0+a_4 x^4+a_8 x^8+a_{12} x^{12}+...\ --- (3)

y_2(x)=\Sigma_{n=0}^{\infty}\ a_{4n+1}\ x^{4n+1}\\ .\ \ \ \ =a_1 x+a_5 x^5+a_9 x^9+a_{13} x^{13}... --(4)\\\\y_1'(x)=\frac{dy_1}{dx}=\Sigma_{n=1}^{\infty}\ 4n*a_{4n}\ x^{4n-1}\\\\y_1''(x)=\frac{d^2y_1}{dx^2}=\Sigma_{n=1}^{\infty}\ 4n(4n-1)*a_{4n}\ x^{4n-2}\ --(5)\\\\y_2'(x)=\frac{dy_2}{dx}=\Sigma_{n=0}^{\infty}\ (4n+1)*a_{4n+1}\ x^{4n}\\\\y_2''(x)=\frac{d^2y_2}{dx^2}=\Sigma_{n=1}^{\infty}\ (4n+1)*4n*a_{4n+1}\ x^{4n-1}\ -- (6)\\\\

Now we use equation 1, to find the coefficients for the two independent solutions.

-x^2*y_1(x)=\Sigma_{n=0}^{\infty}\ (-a_{4n})\ x^{4n+2}=\Sigma_{n=1}^{\infty}\ (-a_{4n-4})\ x^{4n-2}\ --(7)\\\\-x^2*y_2(x)=\Sigma_{n=0}^{\infty}\ (-a_{4n+1})\ x^{4n+3}=\Sigma_{n=1}^{\infty}\ (-a_{4n-3})\ x^{4n-1}\ --(8)

Compare equations (5) with (7) and (6) with (8) we get the coefficients: 

a_{4n}=-\frac{a_{4n-4}}{4n*(4n-1)},n=1,2,3,...\ \ and\\\\ a_{4n+1}=-\frac{a_{4n-3}}{(4n+1)(4n)}


Finally, any linear combination of the two above solutions will be the general solution the given differential equation in (1).

       y(x) = c₁ * y₁(x) + c₂ * y₂(x)

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