What is the solution for 42y^{''}+29y^'-5y=0 ?

The solution for 42y^{''}+29y^'-5y=0 is y=c_{1}e^{t/7}+c_{2}e^{-(5t)/6}

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42y^{''}+29y^'-5y=0

42 y^{^{''}} + 29 y^{^{'}} - 5 y = 0

y = c_{1} e^{\frac{t}{7}} + c_{2} e^{- \frac{5 t}{6}}

Solution steps

42 y^{''} (t) + 29 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{7}} + c_{2} e^{- \frac{5 t}{6}}

y = c_{1} e^{\frac{t}{7}} + c_{2} e^{- \frac{5 t}{6}}

n = 1 \sum \infty \frac{n}{( 3 + n )}

\frac{d}{d x} (ln (3 x^{3} + x + 6))

\frac{d}{d x} (\frac{1 + x ^{2}}{1 + x})

\frac{16}{( 1 - x ) ^{3}}

\int \frac{( 9 - x ^{2} )}{x ^{2}} d x

What is the solution for 42y^{''}+29y^'-5y=0 ?
The solution for 42y^{''}+29y^'-5y=0 is y=c_{1}e^{t/7}+c_{2}e^{-(5t)/6}