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

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

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

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

y = e^{t}

Solution steps

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

Solve linear ODE:

y = e^{t}

y = e^{t}

x \to 0 - lim (ln ∣ e^{x} - 1 ∣)

\frac{d}{d x} (3 x + x^{\frac{1}{2}})

y = (x - 4) (x^{2} + 8), (1, - 27)

\frac{d}{d x} (e^{x} - 1 + e^{x} x)

\frac{\partial}{\partial x} (x^{0.2} y^{0.8})

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