Homogeneous Differential Equations (introduction & example) 

put y= vx and try !!!

bro idk how he got tan in the last one

@@1.8millionvolts87 search for the derivative of the inversa tangent, or arctan.

trig identities @@1.8millionvolts87

son dang nope. You can also use this for x/y by making x=uy, dx=udy+ydu, and u=x/y

Did you not watch the video?

it is like he used dx and dv as variables and moved them to sides at which the respective variable of their derivatives are so we get that and as for integration we need we need a differential variable in the numerator we flip both sides and get the resultant equation for integration

Yeah the constant is G

No, it is not possibly to write every first order ordinary differential equation as an equation of the form dy/dx = G(y/x). This is why these have a special name: homogeneous equations.

Yes, the West likes division a lot.

اختصر ال v مع v cancellation work :D

Mai Korean

excuse me madam, your equation is incomplete!!!! is it equal to .... i mean (X^2-xy + y^2) dx - xy dy = ????

(x^2-xy+y^2)dx-xydy=0 (x^2-xy+y^2)dx=xydy here divide both sides by x^2dx to get 1-y/x+(y^2/x^2)=(y/x)dy/dx now let v=(y/x) and y will be y= vx. to get 1-v+v^2=vdy/dx devide both sides by v (1-v+v^2)/v=dy/dx remember in y=vx, dy/dx= v +xdv/dx since here dy/dx resembles our dy/dx in earlier equation, we equivalent: v+xdv/dx=(1-v+v^2)/v subtract both sides v ((2v^2-v+1)/v=xdv/dx make each side be of the same type i.e v/(2v^2-v+1)dv=(1/x)dx now intergrate both sides [7ln16v^2-8v+8+2arctan{(4vroot7-root7)sq.root7}/7]/28=ln|x|+c then substitute v with (y/x) Am Vincent at University of Nairobi in Kenya. Any problem contact me at kanamwaivincent61@gmail.com

very good👨‍🎓👨‍🎓👨‍🎓

= )

you're welcome

hmmmm, which country?