Question

solve differenciation equation
$ye {}^{ \frac{x}{y} }=(x {e}^{ \frac{x}{y} } + {y}^{2} )$
​

Answer 1

Question:

solve Diffrential equation:

$\\ \displaystyle { \bf \:{ \color{red}{ \longrightarrow ye {}^{ \frac{x}{y} } \: dx=(x {e}^{ \frac{x}{y} } + {y}^{2} ) dy }}}$

Solution:

$\\ \displaystyle { \bf \:{ \dashrightarrow \: ye \: {}^{ \frac{x}{y} } \: dx=(x {e}^{ \frac{x}{y} } + {y}^{2} ) dy }}$

first of all we convert this equation in terms of $\sf \: \frac{ dy}{dx} or \: \frac{dx}{dy}$

now, we get

$\\ \displaystyle { \bf \:{ \dashrightarrow \frac{dx}{dy} = \: \frac{(x {e}^{ \frac{x}{y} } + {{y}^{2} ) }}{\: ye \: {}^{ \frac{x}{y} }} }}$

$\\ \displaystyle { \bf \:{ \dashrightarrow \frac{dx}{dy} = \frac{x}{y} + ye ^{\frac{ - x}{y} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: -(i) }}}$

now by putting $\sf \frac{x}{y} = t$we get,

$\\ \sf \dashrightarrow \: x = y \: t$

now,

$\sf \dashrightarrow \: \frac {dx}{dy} = t + y \: \frac{dt}{dy}$

Substituting the value of dx / dy from equation 1

$\\ \sf \dashrightarrow t+ y e ^{-t} = t + y \frac{dt}{dy}$

$\\ \sf \dashrightarrow dy = e^{t} dt$

now integrating by both side ,

$\\ \sf \dashrightarrow \int dy = \int e^{t} dt$

$\\{\dashrightarrow{\underline{\boxed {\bf \:{ \color{blue} y = e^{t} +c}}}}}\pink\bigstar$

Answer 2

Step-by-step explanation:

Question:

solve Diffrential equation:

$\\ \displaystyle { \bf \:{ \color{red}{ \longrightarrow ye {}^{ \frac{x}{y} } \: dx=(x {e}^{ \frac{x}{y} } + {y}^{2} ) dy }}}$

Solution:

$\\ \displaystyle { \bf \:{ \dashrightarrow \: ye \: {}^{ \frac{x}{y} } \: dx=(x {e}^{ \frac{x}{y} } + {y}^{2} ) dy }}$

first of all we convert this equation in terms of $\sf \: \frac{ dy}{dx} or \: \frac{dx}{dy}$

now, we get

$\\ \displaystyle { \bf \:{ \dashrightarrow \frac{dx}{dy} = \: \frac{(x {e}^{ \frac{x}{y} } + {{y}^{2} ) }}{\: ye \: {}^{ \frac{x}{y} }} }}$

$\\ \displaystyle { \bf \:{ \dashrightarrow \frac{dx}{dy} = \frac{x}{y} + ye ^{\frac{ - x}{y} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: -(i) }}}$

now by putting $\sf \frac{x}{y} = t$we get,

$\\ \sf \dashrightarrow \: x = y \: t$

now,

$\sf \dashrightarrow \: \frac {dx}{dy} = t + y \: \frac{dt}{dy}$

Substituting the value of dx / dy from equation 1

$\\ \sf \dashrightarrow t+ y e ^{-t} = t + y \frac{dt}{dy}$

$\\ \sf \dashrightarrow dy = e^{t} dt$

now integrating by both side ,

$\\ \sf \dashrightarrow \int dy = \int e^{t} dt$

$\\{\dashrightarrow{\underline{\boxed {\bf \:{ \color{blue} y = e^{t} +c}}}}}\pink\bigstar$