Write the function that has derivative y prime = #x^2# and passes through (1,1) ?

2 Answers
Mar 18, 2017

#implies y = 1/3 (x^3 + 2)#

Explanation:

I'd follow the instructions on the packet.

#y' = x^2#

#implies y = x^3/3 + C#

Applying the IV's:

#y(1) = 1 implies 1 = 1/3 + C implies C = 2/3#

So:

#implies y = x^3/3 + 2/3 = 1/3 (x^3 + 2)#

Mar 18, 2017

#y=(x^3)/3+2/3#
#color(white)("XXXXXXXXX")#although, maybe, that's only my opinion.

Explanation:

If #y' = x^2#
then
#color(white)("XXX")y=(x^3)/3+c# for some constant #c#

If #(color(red)x,color(blue)y)=(color(red)1,color(blue)1)# is a solution to this equation
#color(white)("XXX")color(blue)1=color(red)1^3/3+c=1/3+c#

#color(white)("XXXXXX")rarr c=2/3#

and the function is
#color(white)("XXX")y=x^3/3+2/3#