What is the solution for 4y^{''}+28y^'+49y=0,y(0)=1,y^'(0)=-6 ?

The solution for 4y^{''}+28y^'+49y=0,y(0)=1,y^'(0)=-6 is y=e^{-(7t)/2}-(5e^{-(7t)/2}t)/(2)

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4y^{''}+28y^'+49y=0,y(0)=1,y^'(0)=-6

4 y^{^{''}} + 28 y^{^{'}} + 49 y = 0, y (0) = 1, y^{^{'}} (0) = - 6

y = e^{- \frac{7 t}{2}} - \frac{5 e ^{- \frac{7 t}{2}} t}{2}

Solution steps

4 y^{''} (t) + 28 y^{'} (t) + 49 y = 0

Solve linear ODE:

y = e^{- \frac{7 t}{2}} - \frac{5 e ^{- \frac{7 t}{2}} t}{2}

y = e^{- \frac{7 t}{2}} - \frac{5 e ^{- \frac{7 t}{2}} t}{2}

14 y^{^{''}} + 5 y^{^{'}} - 6 y = 0

y^{^{''}} - 24 y^{^{'}} + 10 y = 0

y^{^{''}} = 5 y, y (0) = - 2, y (2) = 10

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 7, y^{^{'}} (0) = 4

y^{^{''}} + 0 y^{^{'}} + 36 y = 0, y (0) = 1

What is the solution for 4y^{''}+28y^'+49y=0,y(0)=1,y^'(0)=-6 ?
The solution for 4y^{''}+28y^'+49y=0,y(0)=1,y^'(0)=-6 is y=e^{-(7t)/2}-(5e^{-(7t)/2}t)/(2)