Math, asked by sk3851610, 6 hours ago

the substitution in dy/dx +y/x = x2 y to make linear equation is

Answers

Answered by Anonymous
0

\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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