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

The solution for y^{''}-y=0,y(0)=e^x is y=c_{1}e^x+c_{2}e^{-x}

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y^{''}-y=0,y(0)=e^x

y^{^{''}} - y = 0, y (0) = e^{x}

y = c_{1} e^{x} + c_{2} e^{- x}

Solution steps

y^{''} (x) - y = 0

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x}

y = c_{1} e^{x} + c_{2} e^{- x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = e^{5 x}

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

\frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} + 10 y = 0

y^{^{''}} + 8 y^{^{'}} - 65 y = 0

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 7 x = 0

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