the substitution in dy/dx +y/x = x2 y to make linear equation is
Answers
\quad\quad\bold{x\:e^{y}=c\:(1-e^{y})}xe
y
=c(1−e
y
)
Step-by-step explanation:
The given differential equation is
\quad\:\:\mathrm{\dfrac{dy}{dx}+\dfrac{1}{x}=\dfrac{e^{y}}{x}}
dx
dy
+
x
1
=
x
e
y
\to \mathrm{\dfrac{dy}{dx}+\dfrac{1}{x}-\dfrac{e^{y}}{x}=0}→
dx
dy
+
x
1
−
x
e
y
=0
\to \mathrm{\dfrac{dy}{dx}+\dfrac{1-e^{y}}{x}=0}→
dx
dy
+
x
1−e
y
=0
\to \mathrm{\dfrac{dy}{1-e^{y}}+\dfrac{dx}{x}=0}→
1−e
y
dy
+
x
dx
=0
\to \mathrm{\dfrac{e^{-y}\:dy}{e^{-y}-1}+\dfrac{dx}{x}=0}→
e
−y
−1
e
−y
dy
+
x
dx
=0
\to \mathrm{-\dfrac{d(e^{-y}-1)}{e^{-y}-1}+d(logx)=0}→−
e
−y
−1
d(e
−y
−1)
+d(logx)=0
Integrating we get
\quad\:\:\mathrm{-\int \dfrac{d(e^{-y}-1)}{e^{-y}-1}+\int d(logx)=0}−∫
e
−y
−1
d(e
−y
−1)
+∫d(logx)=0
\to \mathrm{-log(e^{-y}-1)+logx=logc}→−log(e
−y
−1)+logx=logc
where c is constant of integration
\to \mathrm{log\dfrac{x}{e^{-y}-1}=logc}→log
e
−y
−1
x
=logc
\to \mathrm{\dfrac{x}{e^{-y}-1}=c}→
e
−y
−1
x
=c
\to \mathrm{\dfrac{x\:e^{y}}{1-e^{y}}=c}→
1−e
y
xe
y
=c
\to \boxed{\bold{x\:e^{y}=c\:(1-e^{y})}}→
xe
y
=c(1−e
y
)
This is the required general solution.
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