find the solution of the differential equation ( y^{2}- xy) dx + x^{2} dy = 0
Please help me with this question. Thank you
Hello, Deci!
$\displaystyle \text{Solve: }\:(y^2- xy)\:\!dx + x^2\:\!dy \:=\: 0$
We have: .$\displaystyle \frac{dy}{dx} \:=\:\frac{xy-y^2}{x^2} \quad\Rightarrow\quad \frac{dy}{dx} \:=\:\frac{y}{x} - \left(\frac{y}{x}\right)^2$
Let $\displaystyle v \,=\,\frac{y}{x}\quad\Rightarrow\quad y \:=\:vx\quad\Rightarrow\quad\frac{dy}{dx} \:=\:v + x\frac{dv}{dx}$
Substitute: .$\displaystyle v + x\frac{dv}{dx} \:=\:v - v^2 \quad\Rightarrow\quad x\frac{dv}{dx} \:=\:-v^2 \quad\Rightarrow\quad v^{-2}dv \:=\:-\frac{dx}{x}$
Integrate: .$\displaystyle \int v^{-2}dv \:=\:-\int\frac{dx}{x} \quad\Rightarrow\quad -v^{-1} \:=\:-\ln x + c$
. . . $\displaystyle \frac{1}{v} \:=\:\ln x + c \:=\:\ln x + \ln C \:=\:\ln(Cx)$
. . . $\displaystyle v \:=\:\frac{1}{\ln(Cx)} \quad\Rightarrow\quad \frac{y}{x} \:=\:\frac{1}{\ln(Cx)}$
Therefore: .$\displaystyle y \;=\;\frac{x}{\ln(Cx)}$
Thanks for your help. I have one simple question
<<<< isn't this supposed to be ln x - c or the plus minus sign in c doesn't matter?
and the answer in the ans key is ln x + x/y = C, but when i convert using v = 1 / (ln x + C), then y = vx, so y = x /(ln x + c) >>> ln x - x/y = C , why is it minus?
And is this what ppl call homogenous equation?
Remember that C is an arbitrary constant of integration. In differential equations, you work with these arbitrary constants a lot, so instead of using different letters like:
y=x+C
substitute D=-C, so C=-D
y=x-D
it's typical to just "recycle" the same letter C:
y=x+C
replace C with -C
y=x-C
Since C is an arbitrary constant, these two equations are equivalent.
There is a difference between Soroban's answer and the answer in the book, though: ln x + x/y = C is not the same as ln x - x/y = C, no matter how much you fiddle with C. Soroban's answer is correct.
Homogeneous is used for two different things in differential equations - check the Wikipedia article.
- Hollywood