What is the solution for 38y^{''}-7y= 5/2 ?

The solution for 38y^{''}-7y= 5/2 is y=c_{1}e^{sqrt(7/38)t}+c_{2}e^{-sqrt(7/38)t}-5/14

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38y^{''}-7y= 5/2

38 y^{^{''}} - 7 y = \frac{5}{2}

y = c_{1} e^{\frac{7}{38} t} + c_{2} e^{- \frac{7}{38} t} - \frac{5}{14}

Solution steps

38 y^{''} (t) - 7 y = \frac{5}{2}

Solve linear ODE:

y = c_{1} e^{\frac{7}{38} t} + c_{2} e^{- \frac{7}{38} t} - \frac{5}{14}

y = c_{1} e^{\frac{7}{38} t} + c_{2} e^{- \frac{7}{38} t} - \frac{5}{14}

3 y^{^{''}} + 12 y = - 4 sin (3 t)

y^{^{'''}} + 4 y^{^{'}} = 3 e^{x} (2 x + 5)

y^{^{''''}} - 4 y^{^{''}} = 5 x^{2} - e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 8} e^{2 t}

\frac{d ^{2} x}{d t ^{2}} + k \frac{d x}{d t} + a = 0

What is the solution for 38y^{''}-7y= 5/2 ?
The solution for 38y^{''}-7y= 5/2 is y=c_{1}e^{sqrt(7/38)t}+c_{2}e^{-sqrt(7/38)t}-5/14