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Silimbuga92

Open question

2022-08-17

Solve the given differential equation by using an appropriate substitution.
(x-y)dx+ x dy=0

tkaljegs

Beginner2022-08-18Added 9 answers

The differential equation is given as
(x-y)dx+xdy=0
Given differential equation can be written as

\frac{d y}{d x} = \frac{y - x}{x} = \frac{y}{x} - 1

Let, y=vx. So,

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

Therefore, substituting y=vx in the given differential equation we have

v + x \frac{d v}{d x} = v - 1

\Rightarrow x \frac{d v}{d x} = - 1

\Rightarrow d v = - \frac{d x}{x}

\Rightarrow \int d v = - \int \frac{d x}{x}

, (taking integration both side)

\Rightarrow v = - \ln x + c

, (where c is an integrating constant)

\Rightarrow \frac{y}{x} = - \ln x + c

, (putting

v = \frac{y}{x}

)

\Rightarrow y = x (c - \ln x)

Result:
The solution of the given differential equation is

y = x (c - \ln x)

, where c is an integrating constant.

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