What is the solution for y^{''}+14y^'+49y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+14y^'+49y=0,y(0)=0,y^'(0)=1 is y=e^{-7t}t

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y^{''}+14y^'+49y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = e^{- 7 t} t

Solution steps

y^{''} (t) + 14 y^{'} (t) + 49 y = 0

Solve linear ODE:

y = e^{- 7 t} t

y = e^{- 7 t} t

\frac{\partial}{\partial x} ((2 x + 2 y)^{\frac{1}{2}})

\int \frac{3 ^{x}}{ln ( 3 )} d x

\frac{d}{d x} (sin (2 x^{3}))

6 x y^{^{'}} - 2 y = \frac{( - x ^{2} )}{y ^{5}}

\frac{d}{d x} (\frac{cot ( x )}{x})

What is the solution for y^{''}+14y^'+49y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+14y^'+49y=0,y(0)=0,y^'(0)=1 is y=e^{-7t}t