What is the solution for t^2y^{''}-2ty^'+2y=4t^2 ?

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

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t^2y^{''}-2ty^'+2y=4t^2

t^{2} y^{^{''}} - 2 t y^{^{'}} + 2 y = 4 t^{2}

y = c_{1} t^{2} + c_{2} t + 4 t^{2} ln (t) - 4 t^{2}

Solution steps

t^{2} y^{''} (t) - 2 t y^{'} (t) + 2 y = 4 t^{2}

Solve linear ODE:

y = c_{1} t^{2} + c_{2} t + 4 t^{2} ln (t) - 4 t^{2}

y = c_{1} t^{2} + c_{2} t + 4 t^{2} ln (t) - 4 t^{2}

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