What is the solution for y^{''}-y^'-42y=0 ?

The solution for y^{''}-y^'-42y=0 is y=c_{1}e^{7t}+c_{2}e^{-6t}

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y^{''}-y^'-42y=0

y^{^{''}} - y^{^{'}} - 42 y = 0

y = c_{1} e^{7 t} + c_{2} e^{- 6 t}

Solution steps

y^{''} (t) - y^{'} (t) - 42 y = 0

Solve linear ODE:

y = c_{1} e^{7 t} + c_{2} e^{- 6 t}

y = c_{1} e^{7 t} + c_{2} e^{- 6 t}

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

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\frac{\partial}{\partial y} (x^{\frac{x}{y}} cos (\frac{x}{y}))

((4 - x^{\frac{2}{3}})^{\frac{3}{2}})^{^{'}}

What is the solution for y^{''}-y^'-42y=0 ?
The solution for y^{''}-y^'-42y=0 is y=c_{1}e^{7t}+c_{2}e^{-6t}