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

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

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

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

y = (1 - \frac{1}{2}) e^{\frac{t}{7}} + \frac{1}{2} e^{- \frac{t}{7}}

Solution steps

49 y^{''} (t) - y = 0

Solve linear ODE:

y = (1 - \frac{1}{2}) e^{\frac{t}{7}} + \frac{1}{2} e^{- \frac{t}{7}}

y = (1 - \frac{1}{2}) e^{\frac{t}{7}} + \frac{1}{2} e^{- \frac{t}{7}}

y^{^{''}} + 4 y - 3 y^{^{'}} = 0

- 100 y + 65 y^{^{'}} - 14 y^{^{''}} + y^{^{'''}} = 0

x^{^{''}} + 4 x^{^{'}} = - 2 x^{^{'}} - 8 x + 3 x

12 y^{^{''}} - 5 y - 2 y = 0

y^{^{''}} + 2 y = 0, y (0) = sin (k t)

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