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

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

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

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

y = e^{t}

Solution steps

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

Solve linear ODE:

y = e^{t}

y = e^{t}

(3 cos (t))^{^{'}}

5 x + \frac{7}{x}

\frac{d}{d x} (x^{3} - 9 x + 3)

f (x) = \frac{1}{x + 2}, [3, 6]

y = sin (tan (8 x))

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