What is the solution for t^2y^{''}+9ty^'+7y=0 ?

The solution for t^2y^{''}+9ty^'+7y=0 is y=(c_{1})/t+(c_{2})/(t^7)

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t^2y^{''}+9ty^'+7y=0

t^{2} y^{^{''}} + 9 t y^{^{'}} + 7 y = 0

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{7}}

Solution steps

t^{2} y^{''} (t) + 9 t y^{'} (t) + 7 y = 0

Solve linear ODE:

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{7}}

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{7}}

\int \frac{x}{x ^{2} - 3 x + 2} d x

\int \frac{y ^{2} + 2 y + 1}{( y ^{2} + 1 ) ^{2}} d y

4 y^{^{''}} + 28 y^{^{'}} + 49 y = 0

f (t) = e^{t + 9}

x \to + 1 lim (\frac{4}{x - 1})

What is the solution for t^2y^{''}+9ty^'+7y=0 ?
The solution for t^2y^{''}+9ty^'+7y=0 is y=(c_{1})/t+(c_{2})/(t^7)