What is the solution for t^2y^{''}-3ty^'+4y=t^2*ln(t) ?

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

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t^2y^{''}-3ty^'+4y=t^2*ln(t)

t^{2} y^{^{''}} - 3 t y^{^{'}} + 4 y = t^{2} \cdot ln (t)

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

Solution steps

t^{2} y^{''} (t) - 3 t y^{'} (t) + 4 y = t^{2} \cdot ln (t)

Solve linear ODE:

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

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

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