What is the solution for t^2y^{''}-2y=3t^2-1 ?

The solution for t^2y^{''}-2y=3t^2-1 is y=c_{1}t^2+(c_{2})/t+t^2ln(t)+1/6-(t^2-1)/3

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t^2y^{''}-2y=3t^2-1

t^{2} y^{^{''}} - 2 y = 3 t^{2} - 1

y = c_{1} t^{2} + \frac{c _{2}}{t} + t^{2} ln (t) + \frac{1}{6} - \frac{t ^{2} - 1}{3}

Solution steps

t^{2} y^{''} (t) - 2 y = 3 t^{2} - 1

Solve linear ODE:

y = c_{1} t^{2} + \frac{c _{2}}{t} + t^{2} ln (t) + \frac{1}{6} - \frac{t ^{2} - 1}{3}

y = c_{1} t^{2} + \frac{c _{2}}{t} + t^{2} ln (t) + \frac{1}{6} - \frac{t ^{2} - 1}{3}

x \to 0 lim (\frac{( 1 )}{x})

y^{^{'}} = (1 - y) (2 - y)

(ln (e))^{^{'}}

\frac{d}{d x} (e^{- (x - 1)^{2}})

\frac{d y}{d x} = (4 cos^{2} (y) tan (x))

What is the solution for t^2y^{''}-2y=3t^2-1 ?
The solution for t^2y^{''}-2y=3t^2-1 is y=c_{1}t^2+(c_{2})/t+t^2ln(t)+1/6-(t^2-1)/3