What is the solution for y^{''}+y^'-2y=3t ?

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

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y^{''}+y^'-2y=3t

y^{^{''}} + y^{^{'}} - 2 y = 3 t

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{3 t}{2} - \frac{3}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{3 t}{2} - \frac{3}{4}

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{3 t}{2} - \frac{3}{4}

\frac{d}{d x} (e^{c o s (x^{2})})

\frac{d}{d x} (tan (e^{2 x}) + e^{t a n (2 x)})

x \to - 1 - lim (\frac{1}{( x + 1 ) ^{2}})

\int_{0}^{\infty} 5^{- x} d x

\int sec^{4} (x) d x

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