What is the solution for y^{''}-2y^'+26y=0,y(0)=2,y^'(0)=5 ?

The solution for y^{''}-2y^'+26y=0,y(0)=2,y^'(0)=5 is y=e^t(2cos(5t)+3/5 sin(5t))

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y^{''}-2y^'+26y=0,y(0)=2,y^'(0)=5

y^{^{''}} - 2 y^{^{'}} + 26 y = 0, y (0) = 2, y^{^{'}} (0) = 5

y = e^{t} (2 cos (5 t) + \frac{3}{5} sin (5 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 26 y = 0

Solve linear ODE:

y = e^{t} (2 cos (5 t) + \frac{3}{5} sin (5 t))

y = e^{t} (2 cos (5 t) + \frac{3}{5} sin (5 t))

\frac{\partial}{\partial x} (2 x ln (x^{2} y))

- \frac{2}{s ^{2}}

\int 5 x lo g_{10} (3 - x) d x

x, 0, 1

\frac{d}{d x} (ln (2 x^{3} - x^{2}))

What is the solution for y^{''}-2y^'+26y=0,y(0)=2,y^'(0)=5 ?
The solution for y^{''}-2y^'+26y=0,y(0)=2,y^'(0)=5 is y=e^t(2cos(5t)+3/5 sin(5t))