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Question Number 823 by 112358 last updated on 18/Mar/15

S o l v e t h e f o l l o w i n g d . e b y u s i n g

v = \frac{d y}{d x}, w h e r e v i s a f u n c t i o n o f x .

x^{2} \frac{d^{2} y}{{d x}^{2}} - 2 \frac{d y}{d x} + x = 0

Answered by 123456 last updated on 18/Mar/15

v = \frac{d y}{d x} \Rightarrow \frac{d v}{d x} = \frac{d^{2} y}{{d x}^{2}}

x^{2} \frac{d^{2} y}{{d x}^{2}} - 2 \frac{d y}{d x} + x = 0 \Rightarrow x^{2} \frac{d v}{d x} - 2 v + x = 0

b y t h e c o m e n t

v = - e^{- \frac{2}{x}} \int \frac{e^{\frac{2}{x}}}{x} d x

\frac{d y}{d x} = - e^{- \frac{2}{x}} \int \frac{e^{\frac{2}{x}}}{x} d x

y = \int - e^{- \frac{2}{x}} \int \frac{e^{\frac{2}{x}}}{x} d x d x

y = - \int e^{- \frac{2}{x}} \int \frac{e^{\frac{2}{x}}}{x} d x d x

E i (u) = \int \frac{e^{u}}{u} d u

\int \frac{e^{\frac{2}{x}}}{x} d x

u = \frac{2}{x} \Leftrightarrow x = \frac{2}{u}

d u = - \frac{2 d x}{x^{2}} \Leftrightarrow - \frac{x d u}{2} = \frac{d x}{x} \Leftrightarrow - \frac{d u}{u} = \frac{d x}{x}

- \int \frac{e^{u}}{u} d u = - Ei (u) + C_{1} = - Ei (\frac{2}{x}) + C_{1}

y = \int - e^{- \frac{2}{x}} [- Ei (\frac{2}{x}) + C_{1}] d x

y = \int e^{- \frac{2}{x}} Ei (\frac{2}{x}) d x - C_{1} \int e^{- \frac{2}{x}} d x

Commented by 123456 last updated on 18/Mar/15

x^{2} \frac{d y}{d x} - 2 y + x = 0

y = u v, v \neq 0

\frac{d y}{d x} = \frac{d u}{d x} v + u \frac{d v}{d x}

x^{2} (\frac{d u}{d x} v + u \frac{d v}{d x}) - 2 u v + x = 0

v (x^{2} \frac{d u}{d x} - 2 u) + (x^{2} u \frac{d v}{d x} + x) = 0

seting the red part 0

x^{2} \frac{d u}{d x} - 2 u = 0

x^{2} d u = 2 u d x

\frac{d u}{u} = \frac{2 d x}{x^{2}}

\int \frac{d u}{u} = \int \frac{2 d x}{x^{2}} = 2 \int \frac{d x}{x^{2}}

\ln u = - \frac{2}{x} + A_{1}

u = A_{2} e^{- \frac{2}{x}}

and then

x^{2} u \frac{d v}{d x} + x = 0

x^{2} u d v = - x d x

d v = - \frac{d x}{u x}

\int d v = \int - \frac{d x}{u x} = - \int \frac{d x}{u x}

v = - \frac{1}{A_{2}} \int \frac{e^{\frac{2}{x}}}{x} d x

y = u v = (A_{2} e^{- \frac{2}{x}}) \times (- \frac{1}{A_{2}} \int \frac{e^{\frac{2}{x}}}{x} d x)

y = - e^{- \frac{2}{x}} \int \frac{e^{\frac{2}{x}}}{x} d x

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