(x+2∗y−1)dx+3∗(x+2∗y)dy=0(x+2∗y−1)dx=−3∗(x+2∗y)dyd(x2/2+2∗y∗x−x)=−3∗d(x∗y+y2)∫d(x2/2+2∗y∗x−x)=∫−3∗d(x∗y+y2)x2/2+2∗y∗x−x=−3∗x∗y−3y2+Cx2+4∗y∗x−2∗x+6∗x∗y+6∗y2+C=0x2+10∗y∗x+6∗y2−2∗x+C=0
when x=0, y=2 =>
02+10∗2∗0+6∗22−2∗0+C=024+C=0C=−24
Answer:
x2+10∗y∗x+6∗y2−2∗x−24=0x2+10∗y∗x+6∗y2−2∗x=24
Comments