What is the solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=-1 is y=e^{2t}(cos(t)-3sin(t))

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

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

y = e^{2 t} (cos (t) - 3 sin (t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{2 t} (cos (t) - 3 sin (t))

y = e^{2 t} (cos (t) - 3 sin (t))

\int \frac{x ^{2}}{x ^{3} + 2} d x

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

7

y^{^{'}} + a y + b = 0

(\frac{2 - 2 x}{2 x - x ^{2} + 3})^{^{'}}

What is the solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=-1 is y=e^{2t}(cos(t)-3sin(t))