What is the solution for y^{''}+2y^'-3y=0,y(0)=4,y^'(0)=1 ?

The solution for y^{''}+2y^'-3y=0,y(0)=4,y^'(0)=1 is y=(4-3/4)e^t+3/4 e^{-3t}

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y^{''}+2y^'-3y=0,y(0)=4,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} - 3 y = 0, y (0) = 4, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = \frac{9}{x - 6}

x \to 0 lim (\frac{x}{sin ( 6 x )})

f (x) = sin^{2} (e^{s i n^{2} (x)})

\frac{\partial}{\partial x} (3 y^{6} sin (3 x))

13 sin (x)

What is the solution for y^{''}+2y^'-3y=0,y(0)=4,y^'(0)=1 ?
The solution for y^{''}+2y^'-3y=0,y(0)=4,y^'(0)=1 is y=(4-3/4)e^t+3/4 e^{-3t}