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

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

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

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

x = e^{t} (cos (t) - sin (t))

Solution steps

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

Solve linear ODE:

x = e^{t} (cos (t) - sin (t))

x = e^{t} (cos (t) - sin (t))

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + 49 y = 0, y (π) = 0, y (0) = 0

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

\frac{d ^{2} y}{d x ^{2}} + 5 y = 0

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