# Find a general solution using power series. y' + (3x^2)y = 0 You should assume that the solution to the equation is `y = sum_(n=0)^oo a_n*x^n` , hence, differentiating with respect to x yields:

`y' = sum_(n=1)^oo n*a_n*x^(n-1)`

Multiplying y by `x^2`  yields `sum_(n=0)^oo a_n*x^(n+2), ` hence substituting y by `sum_(n=0)^oo a_n*x^n`  and y' by `sum_(n=1)^oo n*a_n*x^(n-1)`  yields:

`sum_(n=1)^oo n*a_n*x^(n-1) + 3sum_(n=0)^oo a_n*x^(n+2) = 0`

Hence, `sum_(n=1)^oo n*a_n*x^(n-1)=- 3sum_(n=0)^oo a_n*x^(n+2)` You need to form equal power of x, hence you need to substitute the summation indices such that:

`sum_(n=0)^oo (n+3)*a_(n+3)*x^(n+2) =- 3sum_(n=0)^oo a_n*x^(n+2)`

Equating coefficients of like powers yields:

`(n+3)*a_(n+3) = -3a_n`

Hence, `a_(n+3) = -(3a_n)/(n+3)`

You may express the coefficients of solution series such that:

`a_3 = -(3a_0)/(3) ; a_4 = -(3a_1)/(4) ; a_5 = -(3a_2)/(5) ; a_6 = -(3a_3)/(6) = 3^2(a_0)/(3*6) ; a_7 = 3^2(a_1)/(4*7) ; a_8 = -(3a_5)/(8) = 3^2(a_2)/(5*8) ; a_9 = -3^3(a_0)/(3*6*9)`

`a_3 = -(3a_0)/(3) ; a_6 = 3^2(a_0)/(3*6); a_9 = -3^3(a_0)/(3*6*9)...`

`a_(3k+3) = (-1)^(3k+3)*3^(3k+3)*a_0/(3*6*9*...*(3k+3))`

`a_4 = -(3a_1)/(4) ; a_7 = 3^2(a_1)/(4*7) ; ...`

`a_(4k+3) = (-1)^(4k+3)*3^(4k+3)*a_1/(4*7*...*(4k+3))`

`a_(5k+3) = (-1)^(5k+3)*3^(5k+3)*a_2/(5*8*...*(5k+3))`

You may write th general solution to equation:

`y = (-y')/(3x^2)`

`y= -(sum_(n=1)^oo n*a_n*x^(n-1))/(3x^2)`

`y = -(1*a_1/(3x^2) + 2*a_2/(3x) + 3*a_3/3 + 4*a_4*x/3 + ...)`

`y = -(1/3)(a_1/(x^2) + 2*a_2/(x)- 3*(3a_0)/(3))- 4(3a_1)/(4) -... )`

Hence, the general solution to differential equation is `y = -(1/3)(a_1/(x^2) + 2*a_2/(x) - x^(n-3)*(a_(3k+3)sum_(k=0)^oo(-1)^(3k+3)*3^(3k+3)*a_0/(3*6*9*...*(3k+3))+ a_(4k+3)sum_(k=0)^oo(-1)^(4k+3)*3^(4k+3)*a_1/(4*7*...*(4k+3)) + a_(5+3)sum_(k=0)^oo(-1)^(5k+3)*3^(5k+3)*a_2/(5*8*...*(5k+3)))).`

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