What is the solution for y^{''}-2y^'+5y=5,y(0)=2,y^'(0)=1 ?

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

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y^{''}-2y^'+5y=5,y(0)=2,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + 5 y = 5, y (0) = 2, y^{^{'}} (0) = 1

y = e^{t} cos (2 t) + 1

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 5

Solve linear ODE:

y = e^{t} cos (2 t) + 1

y = e^{t} cos (2 t) + 1

t an g e n t f (x) = (\frac{x - 3}{x + 1})^{2}, at x = 0

\int y e^{3 y} d y

\int \frac{sin ( x )}{3} d x

\int_{1}^{2} (x^{2} - 4) d x

\int 2 cos^{2} (t) sin^{2} (t) d t

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