해법
(1−y2)0.5dx=(1+x2)ydy
해법
y=22−c12−arctan2(x)−2c1arctan(x)+(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=−22−c12−arctan2(x)−2c1arctan(x)+(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=22−c12−arctan2(x)−2c1arctan(x)−(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=−22−c12−arctan2(x)−2c1arctan(x)−(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21
솔루션 단계
(1−y2)0.5dx=(1+x2)ydy
분리 가능한 ODE 해결:y=22−c12−arctan2(x)−2c1arctan(x)+(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=−22−c12−arctan2(x)−2c1arctan(x)+(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=22−c12−arctan2(x)−2c1arctan(x)−(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=−22−c12−arctan2(x)−2c1arctan(x)−(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21
y=22−c12−arctan2(x)−2c1arctan(x)+(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=−22−c12−arctan2(x)−2c1arctan(x)+(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=22−c12−arctan2(x)−2c1arctan(x)−(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21,y=−22−c12−arctan2(x)−2c1arctan(x)−(arctan4(x)+4c1arctan3(x)+6c12arctan2(x)+4c13arctan(x)+c14)21