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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

3 V^{^{''}} + 4 V^{^{'}} + 3 V = 0

x^{^{''}} + 4 a^{2} x = 0

q^{^{''}} + 806 q^{^{'}} + 8000 q = 0

y^{^{'''}} - 12 y^{^{''}} + 21 y^{^{'}} + 98 y = 0

y^{^{''}} + 7 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 9

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