What is the solution for y^{''}-2y^'=4t,y(0)=2,y^'(0)=0 ?

The solution for y^{''}-2y^'=4t,y(0)=2,y^'(0)=0 is y=(2-3/2)e^{2t}+3/2-t^2-t

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y^{''}-2y^'=4t,y(0)=2,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} = 4 t, y (0) = 2, y^{^{'}} (0) = 0

y = (2 - \frac{3}{2}) e^{2 t} + \frac{3}{2} - t^{2} - t

Solution steps

y^{''} (t) - 2 y^{'} (t) = 4 t

Solve linear ODE:

y = (2 - \frac{3}{2}) e^{2 t} + \frac{3}{2} - t^{2} - t

y = (2 - \frac{3}{2}) e^{2 t} + \frac{3}{2} - t^{2} - t

\frac{\partial}{\partial x} (- x^{2} + y^{2} - 1)

\frac{d}{d x} ((6 x - x^{2})^{\frac{1}{2}})

\int t cos (t) sin (t) d t

\frac{d}{d x} (\frac{2}{7 x ^{3}})

\frac{d}{d x} (\frac{1}{a ^{2} - x ^{2}})

What is the solution for y^{''}-2y^'=4t,y(0)=2,y^'(0)=0 ?
The solution for y^{''}-2y^'=4t,y(0)=2,y^'(0)=0 is y=(2-3/2)e^{2t}+3/2-t^2-t