What is the solution for y^{''}-12y^'+35y=0,y(0)=e^{kx} ?

The solution for y^{''}-12y^'+35y=0,y(0)=e^{kx} is y=c_{1}e^{7x}+c_{2}e^{5x}

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y^{''}-12y^'+35y=0,y(0)=e^{kx}

y^{^{''}} - 12 y^{^{'}} + 35 y = 0, y (0) = e^{k x}

y = c_{1} e^{7 x} + c_{2} e^{5 x}

Solution steps

y^{''} (x) - 12 y^{'} (x) + 35 y = 0

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} e^{5 x}

y = c_{1} e^{7 x} + c_{2} e^{5 x}

\frac{d ^{3} y}{d x ^{3}} + 5 y = 0

y^{^{''}} + 5 y^{^{'}} - 14 y = 0, y^{^{'}} (0) = 56

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y^{^{'}} (0) = 2, y (0) = 1

y^{^{''}} + 64 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{64}{5}

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

What is the solution for y^{''}-12y^'+35y=0,y(0)=e^{kx} ?
The solution for y^{''}-12y^'+35y=0,y(0)=e^{kx} is y=c_{1}e^{7x}+c_{2}e^{5x}