# Differential equation problems

I need your help to finish my homework. Can somebody help me? I cannot finish all my homework especially in this problem.

1. $\dfrac{dy}{dx} = \dfrac{y^2 -1}{x}$; my answer is $-\dfrac{1}{y} -y = \ln x + x$, is it right?
2. $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$
3. $\dfrac{dy}{dx} = \dfrac{2xy + 3y^2}{x^2 + 2xy}$
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Woah... IMO 1 question/post is a more preferable approach. –  FrenzY DT. Dec 5 '12 at 15:54

Your answer for 1 does not appear to be correct. The equation is separable. Use partial fractions to get something easy to integrate, use properties of logarithms to get the left side into the form $\ln f(y)$, then take $e$ to the power of both sides to eliminate the natural logs.

Problem 3 can be solved using the same method Babak Sorouh used for problem 2. There is a term for a differential equation that can be written in the form $\frac{dy}{dx}=f(\frac yx)$, but I do not recall what it is. In any case, dividing top and bottom by $x^2$ for this problem yields

$$\frac{dy}{dx}=\frac{2(\frac yx)+3(\frac yx)^2}{2(\frac yx)+1}$$

The same substitution $y=xu$ will result in a separable equation.

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thank you very much, really help –  John Dec 5 '12 at 17:19

For 2. The equation can be formed as:$$\frac{dy}{dx}=\frac{x^2\left(1+\frac{y^2}{x^2}\right)}{x^2\left(\frac{y}{x}\right)}$$

then if $x\neq0$ by taking $u=\frac{y}{x}$, we have $$\frac{dy}{dx}=\frac{\left(1+u^2\right)}{u}$$

but $u=\frac{y}{x}$ leads us to $xu=y$ and then $1+u'=y'$ where in $y'=\frac{dy}{dx}$. Now we have $$1+\frac{du}{dx}=\frac{\left(1+u^2\right)}{u}$$ This is a separable ode of first order. We have $$\frac{du}{dx}=\frac{\left(1+u^2\right)}{u}-1=\frac{1-u+u^2}{u}$$ so $$\frac{u\;du}{1-u+u^2}=dx$$ Now take an integral of both sides regarding to corresponding variable. We have $$\frac{1}{2}\ln(1-u+u^2)+\frac{\sqrt{3}}{3}\arctan\left( \frac{\sqrt{3}}{3}(2u-1)\right)=x+C$$ put $u=\frac{y}{x}$.

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