What is the solution for x^2y^{''}+xy^'-y=x^{1/2} ?

The solution for x^2y^{''}+xy^'-y=x^{1/2} is y=c_{1}x+(c_{2})/x-(4x^{1/2})/3

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x^2y^{''}+xy^'-y=x^{1/2}

x^{2} y^{^{''}} + x y^{^{'}} - y = x^{\frac{1}{2}}

y = c_{1} x + \frac{c _{2}}{x} - \frac{4 x ^{\frac{1}{2}}}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} x + \frac{c _{2}}{x} - \frac{4 x ^{\frac{1}{2}}}{3}

y = c_{1} x + \frac{c _{2}}{x} - \frac{4 x ^{\frac{1}{2}}}{3}

x \to \infty lim (\frac{4 ^{x} + 3}{4 ^{x + 1}})

f (x) = - \frac{2}{x}

\int (csc (x)) d x

\frac{d}{d x} (arctan (- 5 x + 1))

f (x) = 3 x^{2} + 5 x - 6

What is the solution for x^2y^{''}+xy^'-y=x^{1/2} ?
The solution for x^2y^{''}+xy^'-y=x^{1/2} is y=c_{1}x+(c_{2})/x-(4x^{1/2})/3