Engineering Maths (solve an equation by separation of variables)

[tmchiam]

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

solve y, by seperation of variables

 

[jayc]

Re: Engineering Maths

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

 

[tmchiam]

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
= ???

 

[jayc]

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 \[n^{a+b} = n^a n^b.\]