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

The solution for y^{''}-2y^'-3y=0,y(0)=b,y^'(0)=2 is y=e^{3t}(b+(2-3b)/4)-(e^{-t}(2-3b))/4

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y^{''}-2y^'-3y=0,y(0)=b,y^'(0)=2

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = b, y^{^{'}} (0) = 2

y = e^{3 t} (b + \frac{2 - 3 b}{4}) - \frac{e ^{- t} ( 2 - 3 b )}{4}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = e^{3 t} (b + \frac{2 - 3 b}{4}) - \frac{e ^{- t} ( 2 - 3 b )}{4}

y = e^{3 t} (b + \frac{2 - 3 b}{4}) - \frac{e ^{- t} ( 2 - 3 b )}{4}

\frac{\partial}{\partial x} (cos (2 (2 y x - 1)))

\frac{\partial}{\partial x} (e^{x yz - 2})

x \to 0 + lim (\frac{2}{x ^{\frac{1}{3}}})

sin^{2} (π x)

\frac{\partial}{\partial x} (98 y^{2} + x^{2} - x^{2} y)

What is the solution for y^{''}-2y^'-3y=0,y(0)=b,y^'(0)=2 ?
The solution for y^{''}-2y^'-3y=0,y(0)=b,y^'(0)=2 is y=e^{3t}(b+(2-3b)/4)-(e^{-t}(2-3b))/4