What is the solution for t^2y^{''}-2y=6t^2-1 ?

The solution for t^2y^{''}-2y=6t^2-1 is y=c_{1}t^2+(c_{2})/t+(12t^2ln(t)-4t^2+3)/6

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t^2y^{''}-2y=6t^2-1

t^{2} y^{^{''}} - 2 y = 6 t^{2} - 1

y = c_{1} t^{2} + \frac{c _{2}}{t} + \frac{12 t ^{2} ln ( t ) - 4 t ^{2} + 3}{6}

Solution steps

t^{2} y^{''} (t) - 2 y = 6 t^{2} - 1

Solve linear ODE:

y = c_{1} t^{2} + \frac{c _{2}}{t} + \frac{12 t ^{2} ln ( t ) - 4 t ^{2} + 3}{6}

y = c_{1} t^{2} + \frac{c _{2}}{t} + \frac{12 t ^{2} ln ( t ) - 4 t ^{2} + 3}{6}

\frac{d}{d y} (y (2 x + y^{3}))

\int 2 sin (x) - tan (x) d x

\frac{\partial}{\partial x} (sin (π (4 x - y)))

\frac{d p}{2 ( p + 3 )} = \frac{d v}{- v}

x \to 0 + lim (x sin (x))

What is the solution for t^2y^{''}-2y=6t^2-1 ?
The solution for t^2y^{''}-2y=6t^2-1 is y=c_{1}t^2+(c_{2})/t+(12t^2ln(t)-4t^2+3)/6