What is the solution for 2y^'+y= 3/2 t^2 ?

The solution for 2y^'+y= 3/2 t^2 is y=(3(t^2-8(t/2-1)))/2+(c_{1})/(e^{t/2)}

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2y^'+y= 3/2 t^2

2 y^{^{'}} + y = \frac{3}{2} t^{2}

y = \frac{3 ( t ^{2} - 8 ( \frac{t}{2} - 1 ) )}{2} + \frac{c _{1}}{e ^{\frac{t}{2}}}

Solution steps

2 y^{'} (t) + y = \frac{3}{2} t^{2}

Solve linear ODE:

y = \frac{3 ( t ^{2} - 8 ( \frac{t}{2} - 1 ) )}{2} + \frac{c _{1}}{e ^{\frac{t}{2}}}

y = \frac{3 ( t ^{2} - 8 ( \frac{t}{2} - 1 ) )}{2} + \frac{c _{1}}{e ^{\frac{t}{2}}}

x \to - 2 lim (\frac{2 x}{x + 2})

\frac{d}{d t} (cos (4 t))

\frac{d}{d x} (11 x^{\frac{1}{2}} + x^{\frac{3}{2}})

\int_{- 6}^{- 2} (- x^{2} - 8 x - 18) d x

(\frac{x}{( x + 1 ) ^{2}})^{^{'}}

What is the solution for 2y^'+y= 3/2 t^2 ?
The solution for 2y^'+y= 3/2 t^2 is y=(3(t^2-8(t/2-1)))/2+(c_{1})/(e^{t/2)}