Question-79966

Question Number 79966 by M±th+et£s last updated on 29/Jan/20

Commented by M±th+et£s last updated on 29/Jan/20

s o l v e t h d O D E

Answered by mr W last updated on 29/Jan/20

\frac{d x}{d y} = \frac{y^{2} + y^{2} e^{{(\frac{x}{y})}^{2}} + 2 x^{2} {(\frac{x}{y})}^{2}}{2 {x y e}^{{(\frac{x}{y})}^{2}}}

\frac{d x}{d y} = \frac{1 + e^{{(\frac{x}{y})}^{2}} + 2 {(\frac{x}{y})}^{4}}{2 (\frac{x}{y}) e^{{(\frac{x}{y})}^{2}}}

l e t u = \frac{x}{y}, i . e . x = u y

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

u + y \frac{d u}{d y} = \frac{1 + e^{u^{2}} + 2 u^{4}}{2 {u e}^{u^{2}}}

y \frac{d u}{d y} = \frac{1 + e^{u^{2}} + 2 u^{4} - 2 u^{2} e^{u^{2}}}{2 {u e}^{u^{2}}}

\frac{2 {u e}^{u^{2}}}{1 + e^{u^{2}} + 2 u^{4} - 2 u^{2} e^{u^{2}}} d u = \frac{d y}{y}

\frac{e^{u^{2}}}{1 + e^{u^{2}} + 2 u^{4} - 2 u^{2} e^{u^{2}}} {d u}^{2} = \frac{d y}{y}

l e t t = u^{2}

\frac{e^{t}}{1 + e^{t} + 2 t^{2} - 2 {t e}^{t}} d t = \frac{d y}{y}

\int \frac{e^{t}}{1 + e^{t} + 2 t^{2} - 2 {t e}^{t}} d t = \int \frac{d y}{y}

\int \frac{e^{t}}{1 + e^{t} + 2 t^{2} - 2 {t e}^{t}} d t = \ln (c y)

\dots . . n o t i n t e g r a b l e \dots

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