What is the solution for 49y^{''}+42y^'+9y=0,y(0)=a,y^'(0)=2 ?

The solution for 49y^{''}+42y^'+9y=0,y(0)=a,y^'(0)=2 is y=ae^{-(3t)/7}+(2+(3a)/7)e^{-(3t)/7}t

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49y^{''}+42y^'+9y=0,y(0)=a,y^'(0)=2

49 y^{^{''}} + 42 y^{^{'}} + 9 y = 0, y (0) = a, y^{^{'}} (0) = 2

y = a e^{- \frac{3 t}{7}} + (2 + \frac{3 a}{7}) e^{- \frac{3 t}{7}} t

Solution steps

49 y^{''} (t) + 42 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = a e^{- \frac{3 t}{7}} + (2 + \frac{3 a}{7}) e^{- \frac{3 t}{7}} t

y = a e^{- \frac{3 t}{7}} + (2 + \frac{3 a}{7}) e^{- \frac{3 t}{7}} t

x = \frac{( y - 3 ) ^{2}}{3}, y = 9 - x

7 x y \frac{d y}{d x} = 2

\frac{d}{d x} (1 + 5 x)

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

12 = \frac{( 33000 ) d y}{d t} + \frac{1}{22 \cdot 1 0 ^{- 6}} y

What is the solution for 49y^{''}+42y^'+9y=0,y(0)=a,y^'(0)=2 ?
The solution for 49y^{''}+42y^'+9y=0,y(0)=a,y^'(0)=2 is y=ae^{-(3t)/7}+(2+(3a)/7)e^{-(3t)/7}t