What is the solution for 28y^{''}-14y^'=0,y(-2)=3,y^'(3)=-1 ?

The solution for 28y^{''}-14y^'=0,y(-2)=3,y^'(3)=-1 is y=-2e^{(t-3)/2}+(3e^{5/2}+2)/(e^{5/2)}

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28y^{''}-14y^'=0,y(-2)=3,y^'(3)=-1

28 y^{^{''}} - 14 y^{^{'}} = 0, y (- 2) = 3, y^{^{'}} (3) = - 1

y = - 2 e^{\frac{t - 3}{2}} + \frac{3 e ^{\frac{5}{2}} + 2}{e ^{\frac{5}{2}}}

Solution steps

28 y^{''} (t) - 14 y^{'} (t) = 0

Solve linear ODE:

y = - 2 e^{\frac{t - 3}{2}} + \frac{3 e ^{\frac{5}{2}} + 2}{e ^{\frac{5}{2}}}

y = - 2 e^{\frac{t - 3}{2}} + \frac{3 e ^{\frac{5}{2}} + 2}{e ^{\frac{5}{2}}}

\int \frac{1}{e ^{x} - 9 e ^{- x}} d x

x \to \infty + lim (tan (x))

x \to 0 lim (\frac{arctan ( x )}{x})

\frac{d}{d x} (\frac{8 x ^{2} + 8 x + 4}{x})

\frac{\partial}{\partial x} (x + 2 y)

What is the solution for 28y^{''}-14y^'=0,y(-2)=3,y^'(3)=-1 ?
The solution for 28y^{''}-14y^'=0,y(-2)=3,y^'(3)=-1 is y=-2e^{(t-3)/2}+(3e^{5/2}+2)/(e^{5/2)}