What is the solution for y^{''}+6y^'+25y=100,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+6y^'+25y=100,y(0)=0,y^'(0)=0 is y=e^{-3t}(-4cos(4t)-3sin(4t))+4

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y^{''}+6y^'+25y=100,y(0)=0,y^'(0)=0

y^{^{''}} + 6 y^{^{'}} + 25 y = 100, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 6 y^{'} (t) + 25 y = 100

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (- e^{x} ln (y))

e^{2 s}

\frac{( x - 1 )}{( x + 1 )}

x \to 4 - lim (\frac{7 x - 1}{x - 4})

\int \frac{x ^{2} + 12 x - 9}{x ^{3} - 9 x} d x

What is the solution for y^{''}+6y^'+25y=100,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+6y^'+25y=100,y(0)=0,y^'(0)=0 is y=e^{-3t}(-4cos(4t)-3sin(4t))+4