What is the solution for y^{''}-4y^'=16t,y(0)=4,y^'(0)=0 ?

The solution for y^{''}-4y^'=16t,y(0)=4,y^'(0)=0 is y=(4-15/4)e^{4t}+15/4-2t^2-t

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y^{''}-4y^'=16t,y(0)=4,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} = 16 t, y (0) = 4, y^{^{'}} (0) = 0

y = (4 - \frac{15}{4}) e^{4 t} + \frac{15}{4} - 2 t^{2} - t

Solution steps

y^{''} (t) - 4 y^{'} (t) = 16 t

Solve linear ODE:

y = (4 - \frac{15}{4}) e^{4 t} + \frac{15}{4} - 2 t^{2} - t

y = (4 - \frac{15}{4}) e^{4 t} + \frac{15}{4} - 2 t^{2} - t

1 - 2 x - 3 x^{2}

x \frac{d y}{d x} = x^{3} + 11 x^{3} y

\frac{tan ( x ) - 1}{sec ( x )}

y y^{^{'}} = 1

\int \frac{e ^{3 x}}{3 x} d x

What is the solution for y^{''}-4y^'=16t,y(0)=4,y^'(0)=0 ?
The solution for y^{''}-4y^'=16t,y(0)=4,y^'(0)=0 is y=(4-15/4)e^{4t}+15/4-2t^2-t