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# Engineering Maths (solve an equation by separation of variables)

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#### tmchiam

##### Newbie level 2
dy/dx = (y-1) / x (x>0)

solve y, by seperation of variables

Re: Engineering Maths

tmchiam said:
dy/dx = (y-1) / x (x>0)

solve y, by seperation of variables

multiply both sides by dx:
dy = (y-1)/x * dx

divide both sides by (y-1):
1/(y-1) dy = 1/x dx

integrate both sides:
ln(y-1) = ln(x)

algebra:
y-1 = x
y = x+1

Engineering Maths

until the integrating parts is correct:
ln(y-1) = ln x + B
where B is an arbitrary constant

pls explain why is it so, for the following solution.
On solving this eqn for y, by first taking the exponential of both sides, we obtain
y = 1 ± e^lnx+B
= 1 ± e^B . e^lnx
= ???

Re: Engineering Maths

oops. You're right. I forgot the constant of integration.

multiply both sides by dx:
$dy = \frac{y-1}{x}dx$

divide both sides by (y-1):
$\frac{1}{y-1} dy = \frac{1}{x} dx$

integrate both sides:
$\ln(y-1) = \ln{x} + C$

Raise everything by exponential:
$y-1 = e^{ln(x) + C} = e^{ln(x)} e^{C}$

algebra:
$y-1 = x e^{C}$ (let's just call $e^{C}$ k since its just a constant anyway)
$y = kx+1$

I think the focus of your question was on the step of getting from $e^{ln(x) + C}$ to $x e^{C}$. It is a property of exponents that $n^{a+b} = n^a n^b.$

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