Question #316644

Substitution as suggested by the equation

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


1
Expert's answer
2022-03-30T11:55:01-0400

(x+2y1)dx+3(x+2y)dy=0(x+2y1)dx=3(x+2y)dyd(x2/2+2yxx)=3d(xy+y2)d(x2/2+2yxx)=3d(xy+y2)x2/2+2yxx=3xy3y2+Cx2+4yx2x+6xy+6y2+C=0x2+10yx+6y22x+C=0(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 =>

02+1020+62220+C=024+C=0C=240^2+10*2*0+6*2^2-2*0+C=0\\ 24+C=0\\ C=-24

Answer:

x2+10yx+6y22x24=0x2+10yx+6y22x=24x^2+10*y*x+6*y^2-2*x-24=0\\ x^2+10*y*x+6*y^2-2*x=24\\


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