Answer
The general solution satisfies the differential equation. The particular solution is $y=-2x + \frac{1}{2} x^3$
Work Step by Step
Find the second derivative of y, Then check to see if it is a solution by substituting $ y $ and $y^{''}$ into the differential equation.
$y=C_1x + C_2x^3$
$y^{'} = C_1 + 3C_2x^2$
$y^{''} = 6C_2x$
Substitute and simplify
$x^2(6C_2x) - 3x(C_1+3C_2x^2) + 3(C_1x + C_2x^3) = 0$
$0=0$, The solution satisfies the differential equation
Solve for the particular solutions
Use the two initial conditions
$ y=0, x=2$ and $y'=4, x=2$
$0=2C_1 +8C_2$
$4=C_1 +12C_2$
Solving for $ C_1 $ and $C_2$ gives
$C_1= -2$
$C_2= \frac{1}{2}$
Resubstitute into y
Particular solution: $y=-2x + \frac{1}{2} x^3$
