What is the solution for u^{''}((x))/x+b^2\H(x)((x))/x =0 ?

The solution for u^{''}((x))/x+b^2\H(x)((x))/x =0 is u(x)=integral of integral of-b^2\H(x)

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u^{''}((x))/x+b^2\H(x)((x))/x =0

u^{^{''}} \frac{( x )}{x} + b^{2} H (x) \frac{( x )}{x} = 0

u (x) = \int \int - b^{2} H (x) d x d x

Solution steps

u^{''} (x) \frac{( x )}{x} + b^{2} H (x) \frac{( x )}{x} = 0

Isolate

u^{''} (x) : u^{''} (x) = - b^{2} H (x)

Solve

u^{''} (x) = - b^{2} H (x) : u^{'} (x) = \int - b^{2} H (x) d x

Solve

u^{'} (x) = \int - b^{2} H (x) d x : u (x) = \int \int - b^{2} H (x) d x d x

u (x) = \int \int - b^{2} H (x) d x d x

(d^{2} - 6 d + 9) y = 0

x^{^{'''}} - 2 x^{^{''}} = 0

(x + 1) y^{^{'}} = y^{2}

y^{^{'}} x^{2} = y

x^{^{'}} = 4 x + 2 e^{t} \cdot x

What is the solution for u^{''}((x))/x+b^2\H(x)((x))/x =0 ?
The solution for u^{''}((x))/x+b^2\H(x)((x))/x =0 is u(x)=integral of integral of-b^2\H(x)