Answer to Question #316644 in Differential Equations for Tobias Felix

Question

Substitution as suggested by the equation

(x + 2y − 1)dx + 3(x + 2y)dy = 0 when y(0) = 2

Answer 1

$(x+2*y-1)dx+3*(x+2*y)dy=0\\ (x+2*y-1)dx=-3*(x+2*y)dy\\ d(x^2/2+2*y*x-x)=-3*d(x*y+y^2)\\ \int d(x^2/2+2*y*x-x)=\int -3*d(x*y+y^2)\\ x^2/2+2*y*x-x=-3*x*y-3y^2+C\\ x^2+4*y*x-2*x+6*x*y+6*y^2+C=0\\ x^2+10*y*x+6*y^2-2*x+C=0\\$

when x=0, y=2 =>

$0^2+10*2*0+6*2^2-2*0+C=0\\ 24+C=0\\ C=-24$

Answer:

$x^2+10*y*x+6*y^2-2*x-24=0\\ x^2+10*y*x+6*y^2-2*x=24\\$