What is the solution for 49y^{''}+28y^'+3y=0,y(2)=9,y^'(2)=3 ?

The solution for 49y^{''}+28y^'+3y=0,y(2)=9,y^'(2)=3 is y=24e^{(-t+2)/7}-15e^{(6-3t)/7}

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49y^{''}+28y^'+3y=0,y(2)=9,y^'(2)=3

49 y^{^{''}} + 28 y^{^{'}} + 3 y = 0, y (2) = 9, y^{^{'}} (2) = 3

y = 24 e^{\frac{- t + 2}{7}} - 15 e^{\frac{6 - 3 t}{7}}

Solution steps

49 y^{''} (t) + 28 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = 24 e^{\frac{- t + 2}{7}} - 15 e^{\frac{6 - 3 t}{7}}

y = 24 e^{\frac{- t + 2}{7}} - 15 e^{\frac{6 - 3 t}{7}}

u^{^{''}} - a H (t) = 0

y^{^{''}} - 8 y + 15 y = 0

(3 + 2) y^{^{''}} - 6 (3 + 3 + 3 + 1) y^{^{'}} + 2 (0 + 7) y = 0

\frac{d ^{2} y}{d x ^{2}} = (- 4 y)

y^{^{''}} - y = 0, y (0) = 0, y (0) = 1

What is the solution for 49y^{''}+28y^'+3y=0,y(2)=9,y^'(2)=3 ?
The solution for 49y^{''}+28y^'+3y=0,y(2)=9,y^'(2)=3 is y=24e^{(-t+2)/7}-15e^{(6-3t)/7}