What is the y^{''}(t)+y^'(t)-2y= 0/y (0)=4,y^'(0)=-5 ?

The y^{''}(t)+y^'(t)-2y= 0/y (0)=4,y^'(0)=-5 is y''(t)+y'(t)-2y=0

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y^{''}(t)+y^'(t)-2y= 0/y (0)=4,y^'(0)=-5

y^{^{''}} (t) + y^{^{'}} (t) - 2 y = \frac{0}{y} (0) = 4, y^{^{'}} (0) = - 5

y'' (t) + y' (t) - 2 y = 0

Solution steps

Find

y^{''} (t) + y^{'} (t) - 2 y = \frac{0}{y} (0) \circ y^{'} (0)

given

y^{''} (t) + y^{'} (t) - 2 y = \frac{0}{y} (0) = 4, y^{'} (0) = - 5

For

y^{''} (t) + y^{'} (t) - 2 y = \frac{0}{y} 0 \circ y^{'} (0)

substitute

y^{'} (0)

with

- 5

y^{''} (t) + y^{'} (t) - 2 y = \frac{0}{y} \cdot 0 \circ (- 5) : y^{''} (t) + y^{'} (t) - 2 y = 0

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

a = x + \frac{1}{x} e n, 1 < x < 5

\frac{3 x + 5}{3 - 2 x} = \frac{5}{3}

\frac{( x ^{2} + 8 x + 16 )}{( x + 4 )} = \frac{- 1}{( x + 2 )}

\frac{x + 1}{x} = \frac{c + 1}{c}

g = \frac{10 m}{s ^{215} m}

What is the y^{''}(t)+y^'(t)-2y= 0/y (0)=4,y^'(0)=-5 ?
The y^{''}(t)+y^'(t)-2y= 0/y (0)=4,y^'(0)=-5 is y''(t)+y'(t)-2y=0