What is the solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-3 ?

The solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-3 is y=e^{-3t}(2cos(4t)+3/4 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-3

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

y = e^{- 3 t} (2 cos (4 t) + \frac{3}{4} sin (4 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 3 t} (2 cos (4 t) + \frac{3}{4} sin (4 t))

y = e^{- 3 t} (2 cos (4 t) + \frac{3}{4} sin (4 t))

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

ln (sec (5 x) + tan (5 x))

\int 5 \frac{cos ( x ) + sin ( 2 x )}{sin ( x )} d x

n = 1 \sum \infty 8 (- \frac{1}{4})^{n}

x \to 0 + lim (2 x ln (x))

What is the solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-3 ?
The solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-3 is y=e^{-3t}(2cos(4t)+3/4 sin(4t))