1. ## first order equations

hi there
i need help in trying to solve the following first order equations .

dy/dx = (x-y) / (x+y) im unsure how to seperatre the variables in this

dy/dx - xy = e^(-x^2)*y^3

(im using ^ as "to the power of ...")

any help appreciated ,thankyou!

2. Hello, philyc86!

The first one is homogeneous . . .

$\frac{dy}{dx} \:=\: \frac{x-y}{x+y}$
Divide top and bottom of the fraction by $x\!: \;\; \frac{dy}{dx} \:=\:\frac{1 - \frac{y}{x}}{1 + \frac{y}{z}}$

Let: . $v \:=\: \frac{y}{x} \quad\Rightarrow\quad y \:=\: vx \quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}$

Substitute: . $v + x\frac{dv}{dx} \:=\:\frac{1-v}{1+v} \quad\Rightarrow\quad x\frac{dv}{dx} \:=\:\frac{1-v}{1+v} - v$

. . $x\frac{dv}{dx} \:=\:\frac{1-2v-v^2}{1+v} \quad\Rightarrow\quad x\frac{dv}{dx} \:=\:-\frac{v^2+2v-1}{v+1}dx$

Separate variables: . $\frac{v+1}{v^2+2x-1}dv \:=\:-\frac{dx}{x}$

Integrate: . $\tfrac{1}{2}\ln(v^2+2v-1) \:=\:-\ln x + c_1 \quad\Rightarrow\quad \tfrac{1}{2}\ln(v^2+2v-1) \:=\:-\ln x + \ln c_2$

. . $\ln(v^2+2v-1) \;\;=\;\;-2\ln x + \ln C \;\;=\;\;\ln(x^{-2}) + \ln C \;\;=\;\;\ln(Cx^{-2})$

Then: . $v^2 + 2v - 1 \:=\:Cx^{-2}$

Back-substitute: . $\left(\frac{y}{x}\right)^2 + 2\left(\frac{y}{x}\right) - 1 \:=\:Cx^{-2}$

. . Multiply by $x^2\!:\;\;\boxed{y^2 + 2xy - x^2 \:=\:C}$

3. Hi,
Originally Posted by philyc86
dy/dx = (x-y) / (x+y)
Another approach :

Substitute $u=x+y$ (so $y=u-x$ and $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u} {\mathrm{d}x}-1$) :

$\frac{\mathrm{d}u}{\mathrm{d}x}-1=\frac{x-(u-x)}{u}\implies\frac{\mathrm{d}u}{\mathrm{d}x}=\fra c{2x-u+u}{u}\implies\frac{\mathrm{d}u}{\mathrm{d}x}=\fr ac{2x}{u}\implies u\,\mathrm{d}u=2x\,\mathrm{d}x$

Now integrate both sides to get $\frac{u^2}{2}=x^2+C\implies u^2=2x^2+C'$. Then back-substitute and you'll find Soroban's answer : $y^2+2xy-x^2=C'$.