Lili W.
asked • 03/04/22
Dayv O. answered • 03/04/22
Caring Super Enthusiastic Knowledgeable Calculus Tutor
have after integration
ln(y-4)=(1/3)x3+C
if when x=0 y=-7
have ln(-11)=C
and ln is not defined for 0 and negative values
update: see my comments and answer by Yefim
since it is known that y(0)= -7, know y<4
therefore ln|y-4| can be written ln(4-y)
so answer is ln(4-y)=(1/3)x3+ln11
check, when x=0 y= -7
Yefim S. answered • 03/04/22
Math Tutor with Experience
By integration lnIy - 4I = x3/3 + C; lnI- 7 - 4I = C; C = ln11; lnIy - 4I = x3/3 + ln11.
Dayv O.
It is then two functions y(x) are created, because of the absolute value application. y1=11e^(x^3/3)+4 for y>4 and y2=4-11e^(x^3/3) for y<4, y=4 is not valid. So for x=1, y equals either 11e^(1/3)+4 or 4-11e^(1/3).03/04/22
Dayv O.
That is y(x) provided in your solution is not a function. Answer should be ln(4-y)=[(1/3)x^3]+ln11.03/05/22
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Dayv O.
if the variable in the ln function is allowed to take on negative values then absolute value must be applied to variable just as Yefim did. Note then that two functions are specified. In this case y1=11e^(x^3/3)+4 for y>4 and y2=4-11e^(x^3/3) for y<403/04/22