Question

The solution of dy/dx = (x - y)^2 , y(0) = 0

Answer 1

Answer:

$(1-x+y)=(1+x-y)e^{2x}$

Step-by-step explanation:

$\frac{dy}{dx}=(x-y)^2\\\\let \;x-y=u\\\\1-\frac{dy}{dx}=\frac{du}{dx}\\\\1-\frac{du}{dx}=u^2\\\\\frac{du}{dx}=1-u^2\\\\\frac{du}{1-u^2}=dx\\\\\int\ \frac{1}{1-u^2}\,du=\int \,dx\\\\\frac{1}{2}log\frac{1-u}{1+u}=x+c\\\\given\; it\; passes\; through\; (0,0)\\for \;(x,y)\\\\\frac{1}{2}log\frac{1-x+y}{1+x-y}=x+c\\\\\frac{1}{2}log1=0+c\\\\c=0\\\\log\frac{1-x+y}{1+x-y}=2x\\\\(1-x+y)=(1+x-y)e^{2x}$

Hope it Helps