What is the solution for x^2y^{''}+3xy^'-15y=0 ?

The solution for x^2y^{''}+3xy^'-15y=0 is y=c_{1}x^3+(c_{2})/(x^5)

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x^2y^{''}+3xy^'-15y=0

x^{2} y^{^{''}} + 3 x y^{^{'}} - 15 y = 0

y = c_{1} x^{3} + \frac{c _{2}}{x ^{5}}

Solution steps

x^{2} y^{''} (x) + 3 x y^{'} (x) - 15 y = 0

Solve linear ODE:

y = c_{1} x^{3} + \frac{c _{2}}{x ^{5}}

y = c_{1} x^{3} + \frac{c _{2}}{x ^{5}}

in v erse l a pl a ce \frac{6}{s ^{2} + 5 s + 6}

\frac{\partial}{\partial x} (2 x e^{y})

t an g e n t \frac{1}{x}, at x = 3

\int \frac{1}{x sin ^{2} ( ln ( x ) )} d x

\int_{0}^{5} \frac{5 t}{( t - 6 ) ^{2}} d t

What is the solution for x^2y^{''}+3xy^'-15y=0 ?
The solution for x^2y^{''}+3xy^'-15y=0 is y=c_{1}x^3+(c_{2})/(x^5)