Mathematics

# The solution of $e^x\sqrt{1-y^2}dx+\dfrac{y}{x}dy=0$.

##### SOLUTION
${ e }^{ x }\sqrt { 1-{ y }^{ 2 } } dx=\cfrac { -y }{ x } dy$
On separating the variables and integrating, we get
$\int { { e }^{ x }xdx } =\int { \cfrac { -y }{ \sqrt { 1-{ y }^{ 2 } } } dy }$
For RHS, let $I=\int { \cfrac { -y }{ \sqrt { 1-{ y }^{ 2 } } } dy }$
Let $\sqrt { 1-{ y }^{ 2 } } =z$
So, $\cfrac { -2y }{ \sqrt { 1-{ y }^{ 2 } } } dy=dz$
So, $\cfrac { -y }{ \sqrt { 1-{ y }^{ 2 } } } dy=dz$
So, $z=\int { dz } =z+c=\sqrt { 1-{ y }^{ 2 } } +c$
For RHS,
Let $J=\int { x{ e }^{ x }dx }$
Using product rule, we get
$J=x\int { { e }^{ x }dx } -\int { \left[ \cfrac { d\left( x \right) }{ dx } \int { { e }^{ x }dx } \right] dx }$
$J+x{ e }^{ x }-{ e }^{ x }={ e }^{ x }\left( x-1 \right) +c$
So, going to original equation, we get
${ e }^{ x }\left( x-1 \right) =\sqrt { 1-{ y }^{ 2 } } +c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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