What is the solution for 9y^{''}-5y^'=0,y^1(0)=e^{-7x} ?

The solution for 9y^{''}-5y^'=0,y^1(0)=e^{-7x} is y=c_{1}e^{(5x)/9}+c_{2}

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9y^{''}-5y^'=0,y^1(0)=e^{-7x}

9 y^{^{''}} - 5 y^{^{'}} = 0, y^{1} (0) = e^{- 7 x}

y = c_{1} e^{\frac{5 x}{9}} + c_{2}

Solution steps

9 y^{''} (x) - 5 y^{'} (x) = 0

Solve linear ODE:

y = c_{1} e^{\frac{5 x}{9}} + c_{2}

y = c_{1} e^{\frac{5 x}{9}} + c_{2}

x^{^{''}} + 5 \cdot x = 0, x (1) = 0, x (0) = 0

(D^{4} - 1) y = 0

y^{^{''}} - 4 y^{^{'}} = 0, y (0) = 5 - 5 e^{4}, y (1) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + 2 y = 0

What is the solution for 9y^{''}-5y^'=0,y^1(0)=e^{-7x} ?
The solution for 9y^{''}-5y^'=0,y^1(0)=e^{-7x} is y=c_{1}e^{(5x)/9}+c_{2}