What is the solution for x^2y^{''}+7xy^'+8y=0,y(1)=0,y^'(1)=-5 ?

The solution for x^2y^{''}+7xy^'+8y=0,y(1)=0,y^'(1)=-5 is y=-5/(2x^2)+5/(2x^4)

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x^2y^{''}+7xy^'+8y=0,y(1)=0,y^'(1)=-5

x^{2} y^{^{''}} + 7 x y^{^{'}} + 8 y = 0, y (1) = 0, y^{^{'}} (1) = - 5

y = - \frac{5}{2 x ^{2}} + \frac{5}{2 x ^{4}}

Solution steps

x^{2} y^{''} (x) + 7 x y^{'} (x) + 8 y = 0

Solve linear ODE:

y = - \frac{5}{2 x ^{2}} + \frac{5}{2 x ^{4}}

y = - \frac{5}{2 x ^{2}} + \frac{5}{2 x ^{4}}

\frac{\partial}{\partial y} (ln (2 x + 3 y))

\frac{d}{d x} (\frac{x ^{3}}{2} + \frac{1}{6 x})

x^{2} + 2 x - 6

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

\frac{d}{d t} (2 t - 3)

What is the solution for x^2y^{''}+7xy^'+8y=0,y(1)=0,y^'(1)=-5 ?
The solution for x^2y^{''}+7xy^'+8y=0,y(1)=0,y^'(1)=-5 is y=-5/(2x^2)+5/(2x^4)