What is a solution to the differential equation #dy/dx=y^2/x^2+y/x+1#?
thus, plugging this into the original.....
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The differential equation ( \frac{dy}{dx} = \frac{y^2}{x^2} + \frac{y}{x} + 1 ) can be solved using an integrating factor.
First, rewrite the equation in a standard form:
[ \frac{dy}{dx} = \frac{y^2}{x^2} + \frac{y}{x} + 1 ]
[ \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2} + 1 ]
The integrating factor ( \mu(x) ) is given by ( \mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln{x}} = \frac{1}{x} ).
Multiply both sides of the equation by ( \frac{1}{x} ):
[ \frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = \frac{y^2}{x^3} + \frac{1}{x} ]
Now, rewrite the left-hand side as the derivative of ( \left( \frac{y}{x} \right) ):
[ \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{y^2}{x^3} + \frac{1}{x} ]
Integrate both sides with respect to ( x ):
[ \int \frac{d}{dx} \left( \frac{y}{x} \right) dx = \int \left( \frac{y^2}{x^3} + \frac{1}{x} \right) dx ]
[ \frac{y}{x} = -\frac{y^2}{2x^2} + \ln{|x|} + C ]
[ y = -\frac{xy^2}{2x^2} + x \ln{|x|} + Cx ]
[ y = -\frac{y^2}{2x} + x \ln{|x|} + Cx ]
This is the general solution to the differential equation.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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