Answer:
Explanation:
To find the value of y at the point where x=1, we can substitute x=1 into the equation:
x^2+xy+y^3=1
1+y+y^3=1
y+y^3=0
y(y^2+1)=0
y=0 or y=-1 or y=1
To find the value of y' at the point where x=1, we need to take the derivative of the equation with respect to x:
2x + y + xy' + 3y^2y' = 0
y' = -(2x+y)/(3y^2+x)
When x = 1, y' = -(2(1)+y)/(3y^2+1)
To find the value of y'' at the point where x=1, we need to take the derivative of the equation with respect to x again:
2 + y' + xy'' + 3(2y)(y') = 0
y'' = -(2+y')/(3(2y)+(x))
When x = 1, y'' = -(2+y')/(3(2y)+(1))
To find the value of y''' at the point where x=1, we need to take the derivative of the equation with respect to x one more time:
y'' + xy''' + 3(2y')(y') + 3y(3y)(y') = 0
y''' = -(y'' + 3(2y')(y') + 3y(3y)(y'))/(x)
When x = 1, y''' = -(y'' + 3(2y')(y') + 3y(3y)(y'))/(1)
It is important to note that to find the exact value of y, y', y'', y''' we need to find the value of y in the equation x^2+xy+y^3=1 and substitute it into the derivative equations, but without the equation solved we can't find the exact values.