Resolved: "Solve the equation 2yxdydx=ln⁡y2−ln⁡x2"

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babeeb0oL

Answered question

2021-09-16

Solve the equation
$\frac{2 y}{x} \frac{d y}{d x} = \ln y^{2} - \ln x^{2}$

Answer & Explanation

doplovif

Skilled2021-09-17Added 71 answers

Take the integral:
$\int (\log (y^{2}) - \log (x^{2})) d x$
Integrate the sum term by term and factor out constants:
$= \log (y^{2}) \int 1 d x - \int \log (x^{2}) d x$
The integral of 1 is x:
$= x \log (y^{2}) - \int l \log (x^{2}) d x$
Rewrite the integrand: $\log (x^{2}) = 2 \log (x)$ :
$= x \log (y^{2}) - \int 2 \log (x) d x$
Factor out constants:
$= x \log (y^{2}) - 2 \int \log (x) d x$
For the integrand log(x), integrate by parts, $\int f d g = f g - \int g d f$ , where
$f = \log (x), d g = d x, d f = \frac{1}{x} d x, g = x$ :
$= - 2 x \log (x) + x \log (y^{2}) + 2 \int 1 d x$
The int of 1 is x:
$= x \log (y^{2}) + 2 x - 2 x \log (x) + constant$
Factor the answer a different way:
$= x (- 2 \log (x) + \log (y^{2}) + 2) + constant$
Which is equivalent for restricted x and y values to:
Answer
$= x (- \log (x^{2}) + \log (y^{2}) + 2) + constan t$

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