Get the answer to "Solve it pls xy′−y=2xln⁡x"

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stop2dance3l

Answered question

2021-12-30

Solve it pls $x y^{'} - y = 2 x \ln x$

Answer & Explanation

eninsala06

Beginner2021-12-31Added 37 answers

We need to solve the differential equation:
$x y^{'} - y = 2 x \ln x$
$x y^{'} - y = 2 x \ln x$
$y^{'} - \frac{y}{x} = 2 \ln x$ (1)
Integrating factor $= e^{- \int \frac{1}{x} d x} = e^{- \ln x} = \frac{1}{x}$
Multiplying equation (1) throughout an integrating factor we get:
$\frac{1}{x} y^{'} - \frac{y}{x^{2}} = \frac{2 \ln x}{x}$
$\frac{d}{d x} (y \frac{1}{x}) = \frac{2 \ln x}{x}$
$y \cdot \frac{1}{x} = \int \frac{2 \ln x}{x} d x$
Substitute $\ln x = t$ we get:
$y \cdot \frac{1}{x} = t^{2} + C$
Replace t by ln x
$y \cdot \frac{1}{x} = {(\ln x)}^{2} + C$
$y (x) = x {(\ln x)}^{2} + C x$

soanooooo40

Beginner2022-01-01Added 35 answers

First we put the equation in standard form:

x y^{'} - y = 2 x \ln x \Rightarrow y^{'} - \frac{y}{x} = 2 \ln x

Thus

P (x) = \frac{- 1}{x}, Q (x) = 2 \ln x

The integrating factor is:

μ (x) = e^{\int \frac{- 1}{x} d x} = e^{- \ln x} = \frac{1}{x}

Next we multiply both sides of the standard form by

μ (x)

and integrate:

\frac{1}{x} y^{'} + \frac{y}{x^{2}} = \frac{2}{x} \ln x

\Rightarrow \frac{d}{d x} [\frac{1}{x} \cdot y] = \frac{2}{x} \ln x

\Rightarrow y = X [{(\ln x)}^{2} + C] = x {(\ln x)}^{2} + x C

user_27qwe

Skilled2022-01-05Added 375 answers

\(\begin{array}{}xy-y=2x

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