What is the solution for y^{''}+ky=t^2e^{3t},y(0)=1,y^'(0)=-1 ?

The solution for y^{''}+ky=t^2e^{3t},y(0)=1,y^'(0)=-1 is

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y^{''}+ky=t^2e^{3t},y(0)=1,y^'(0)=-1

y^{^{''}} + k y = t^{2} e^{3 t}, y (0) = 1, y^{^{'}} (0) = - 1

y = cos (k t) (1 + \frac{2 k - 54}{( k + 9 ) ( k ^{2} + 18 k + 81 )}) + \frac{sin ( k t ) ( - k ^{3} - 27 k ^{2} - 225 k - 783 )}{k ( k + 9 ) ^{3}} + \frac{e ^{3 t} t ^{2}}{9 + k} - \frac{12 e ^{3 t} t}{( k + 9 ) ^{2}} - \frac{e ^{3 t} ( 2 k - 54 )}{( k ^{2} + 18 k + 81 ) ( 9 + k )}

Solution steps

y^{''} (t) + k y (t) = t^{2} e^{3 t}

Solve linear ODE:

y = cos (k t) (1 + \frac{2 k - 54}{( k + 9 ) ( k ^{2} + 18 k + 81 )}) + \frac{sin ( k t ) ( - k ^{3} - 27 k ^{2} - 225 k - 783 )}{k ( k + 9 ) ^{3}} + \frac{e ^{3 t} t ^{2}}{9 + k} - \frac{12 e ^{3 t} t}{( k + 9 ) ^{2}} - \frac{e ^{3 t} ( 2 k - 54 )}{( k ^{2} + 18 k + 81 ) ( 9 + k )}

y = cos (k t) (1 + \frac{2 k - 54}{( k + 9 ) ( k ^{2} + 18 k + 81 )}) + \frac{sin ( k t ) ( - k ^{3} - 27 k ^{2} - 225 k - 783 )}{k ( k + 9 ) ^{3}} + \frac{e ^{3 t} t ^{2}}{9 + k} - \frac{12 e ^{3 t} t}{( k + 9 ) ^{2}} - \frac{e ^{3 t} ( 2 k - 54 )}{( k ^{2} + 18 k + 81 ) ( 9 + k )}

y^{^{''}} + 2 y^{^{'}} + y = t^{2}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 4 y = g (t), y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 6

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 5

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 x + 4

What is the solution for y^{''}+ky=t^2e^{3t},y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+ky=t^2e^{3t},y(0)=1,y^'(0)=-1 is