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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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