What is the solution for y^{''}+(14)/t y^'+(42)/(t^2)y=0 ?

The solution for y^{''}+(14)/t y^'+(42)/(t^2)y=0 is y=(c_{1})/(t^6)+(c_{2})/(t^7)

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y^{''}+(14)/t y^'+(42)/(t^2)y=0

y^{^{''}} + \frac{14}{t} y^{^{'}} + \frac{42}{t ^{2}} y = 0

y = \frac{c _{1}}{t ^{6}} + \frac{c _{2}}{t ^{7}}

Solution steps

y^{''} (t) + \frac{14}{t} y^{'} (t) + \frac{42}{t ^{2}} y = 0

Solve linear ODE:

y = \frac{c _{1}}{t ^{6}} + \frac{c _{2}}{t ^{7}}

y = \frac{c _{1}}{t ^{6}} + \frac{c _{2}}{t ^{7}}

\frac{d ^{2}}{d x ^{2}} (3^{x})

\frac{d}{d x} (sin (e^{x^{2}}) \cdot cos (x))

\int \frac{1}{x ^{3} - 3 x ^{2}} d x

\frac{d}{d x} (x^{2} - 3 x + 12)

f (x) = 2 sin (x), at x = \frac{3 π}{4}

What is the solution for y^{''}+(14)/t y^'+(42)/(t^2)y=0 ?
The solution for y^{''}+(14)/t y^'+(42)/(t^2)y=0 is y=(c_{1})/(t^6)+(c_{2})/(t^7)