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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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