What is the solution for x^2y^{''}-5xy^'-7y=0 ?

The solution for x^2y^{''}-5xy^'-7y=0 is y=c_{1}x^7+(c_{2})/x

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x^2y^{''}-5xy^'-7y=0

x^{2} y^{^{''}} - 5 x y^{^{'}} - 7 y = 0

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

Solution steps

x^{2} y^{''} (x) - 5 x y^{'} (x) - 7 y = 0

Solve linear ODE:

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

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

y^{^{'}} = - 5 cos (3 x) y

f (x) = e^{- 6 x}

\frac{\partial}{\partial x} (x^{9} - 7 y)

\int e^{1.2 x} d x

4 - x + 16

What is the solution for x^2y^{''}-5xy^'-7y=0 ?
The solution for x^2y^{''}-5xy^'-7y=0 is y=c_{1}x^7+(c_{2})/x