Solving differential equations with negative exponents

ucci

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Apr 29, 2008
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Find an equation of the curve that passes through the point (1,1)
(dy/dx)= (y^2/x^3)

This what I've done so far and also where I get stuck. please help.
DX (1/ X^3) = DY(1/Y^2)
DX X^-3= DY Y^-2
(-1/2) X^-2= (-Y)^ -1
 
You could separate variables and get \(\displaystyle \frac{dy}{y^{2}}=\frac{dx}{x^{3}}\)

Integrate: \(\displaystyle \int\frac{1}{y^{2}}dy=\int\frac{1}{x^{3}}dx\)

\(\displaystyle \frac{-1}{y}=\frac{-1}{2x^{2}}+C\)

\(\displaystyle y=\frac{-2x^{2}}{2Cx^{2}-1}\)

Now, solve for C by using the (1,1) and you have it kicked.
 
ucci said:
Find an equation of the curve that passes through the point (1,1)
(dy/dx)= (y^2/x^3)

This what I've done so far and also where I get stuck. please help.
DX (1/ X^3) = DY(1/Y^2)
DX X^-3= DY Y^-2
(-1/2) X^-2= (-Y)^ -1
You have done everything okay - except you forgot your constant of integration
 
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