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November 28th, 2016, 06:11 AM   #1
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help separate eq

xy'-y=(x+y)(ln(x+y)-ln(x))

please help just need the start
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November 28th, 2016, 07:34 AM   #2
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Substitute y = xu.
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November 29th, 2016, 07:52 PM   #3
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Originally Posted by ibanez1608 View Post
xy'-y=(x+y)(ln(x+y)-ln(x))

please help just need the start
If you can write the equation as $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = f\left( \frac{y}{x} \right) \end{align*}$ then the substitution $\displaystyle \begin{align*} u = \frac{y}{x} \end{align*}$ is appropriate...

$\displaystyle \begin{align*} x\,\frac{\mathrm{d}y}{\mathrm{d}x} - y &= \left( x + y \right) \left[ \ln{ \left( x + y \right) } - \ln{ \left( x \right) } \right] \\ x\,\frac{\mathrm{d}y}{\mathrm{d}x} - y &= \left( x + y \right) \ln{ \left( \frac{x + y}{x} \right) } \\ x\,\frac{\mathrm{d}y}{\mathrm{d}x} - y &= \left( x + y \right) \ln{ \left( 1 + \frac{y}{x} \right) } \\ x\,\frac{\mathrm{d}y}{\mathrm{d}x} &= \left( x + y \right) \ln{ \left( 1 + \frac{y}{x} \right) } + y \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \left( \frac{x + y}{x} \right) \ln{ \left( 1 + \frac{y}{x} \right) } + \frac{y}{x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \left( 1 + \frac{y}{x} \right) \ln{ \left( 1 + \frac{y}{x} \right) } + \frac{y}{x} \end{align*}$

So now substitute $\displaystyle \begin{align*} u = \frac{y}{x} \implies y = u\,x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = u + x\,\frac{\mathrm{d}u}{\mathrm{d}x} \end{align*}$ giving

$\displaystyle \begin{align*} u + x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \left( 1 + u \right) \ln{ \left( 1 + u \right) } + u \\ x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \left( 1 + u \right) \ln{ \left( 1 + u \right) } \\ \frac{1}{\left( 1 + u \right) \ln{ \left( 1 + u \right) }}\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{x} \\ \int{ \frac{1}{\left( 1 + u \right) \ln{ \left( 1 + u \right) }}\,\frac{\mathrm{d}u}{\mathrm{d}x} \,\mathrm{d}x } &= \int{ \frac{1}{x}\,\mathrm{d}x } \\ \int{ \frac{1}{\left( 1 + u \right) \ln{ \left( 1 + u \right) }} \,\mathrm{d}u } &= \ln{ \left| x \right| } + C_1 \\ \int{ \frac{1}{v}\,\mathrm{d}v } &= \ln{ \left| x \right| } + C_1 \textrm{ when substituting } v = \ln{ \left( 1 + u \right) } \implies \mathrm{d}v = \frac{1}{1 + u}\,\mathrm{d}u \\ \ln{ \left| v \right| } + C_2 &= \ln{ \left| x \right| } + C_1 \\ \ln{ \left| \ln{ \left( 1 + u \right) } \right| } &= \ln{ \left| x \right| } + C \textrm{ where } C = C_1 - C_2 \\ \ln{ \left| \ln{ \left( 1 + \frac{y}{x} \right) } \right| } &= \ln{ \left| x \right| } + C \\ \ln{ \left| \ln{ \left( \frac{x + y}{x} \right) } \right| } &= \ln{ \left| x \right| } + C \\ \ln{ \left| \ln{\left( x + y \right) } - \ln{ \left( x \right) } \right| } &= \ln{ \left| x \right| } + C \\ \ln{ \left| \ln{\left( x + y \right) } - \ln{ \left( x \right) } \right| } - \ln{ \left| x \right| } &= C \\ \ln{ \left| \frac{\ln{\left( x + y \right) } - \ln{\left( x \right) }}{x} \right| } &= C \end{align*}$
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