What is the solution for 3y^'-2y=e^{-pi/2} ?

The solution for 3y^'-2y=e^{-pi/2} is y=(e^{(2(2t+c_{1})-3pi)/6})/(2)-1/(2e^{pi/2)}

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3y^'-2y=e^{-pi/2}

3 y^{^{'}} - 2 y = e^{- \frac{π}{2}}

y = \frac{e ^{\frac{2 ( 2 t + c _{1} ) - 3 π}{6}}}{2} - \frac{1}{2 e ^{\frac{π}{2}}}

Solution steps

3 y^{'} (t) - 2 y = e^{- \frac{π}{2}}

Solve separable ODE:

y = \frac{e ^{\frac{2 ( 2 t + c _{1} ) - 3 π}{6}}}{2} - \frac{1}{2 e ^{\frac{π}{2}}}

y = \frac{e ^{\frac{2 ( 2 t + c _{1} ) - 3 π}{6}}}{2} - \frac{1}{2 e ^{\frac{π}{2}}}

y = x^{2 c o s (x)}

\frac{d}{d x} (3 - ∣ x - 3 ∣)

n = 0 \sum \infty 7 (- \frac{1}{8})^{n}

x y^{^{'}} + 4 y = 2 x^{3}

y = 7 x + 2, (2, 4)

What is the solution for 3y^'-2y=e^{-pi/2} ?
The solution for 3y^'-2y=e^{-pi/2} is y=(e^{(2(2t+c_{1})-3pi)/6})/(2)-1/(2e^{pi/2)}