Asked by bobpursley
dy/dx = 2y^2
Integrating...y=2/3 y^3 + C
put 1,-1 into the equation, and solve for C.
Then find the y for x=2
if y= a^uhttp://math2.org/math/integrals/tableof.htm
see exponential functions.
dy/dx=2y^2 and if y=-1 when x=1, then when x=2, y=?
how do i get x in the equation. do i integrate dy/dx because that would be (2/3)y^3 +C.
2)how i would integrate 13^x-11^x.
lets say i integrate 13^x, would that be 13^x*log 13. i know there is a formula for that but i don't remember
Integrating...y=2/3 y^3 + C
put 1,-1 into the equation, and solve for C.
Then find the y for x=2
if y= a^uhttp://math2.org/math/integrals/tableof.htm
see exponential functions.
dy/dx=2y^2 and if y=-1 when x=1, then when x=2, y=?
how do i get x in the equation. do i integrate dy/dx because that would be (2/3)y^3 +C.
2)how i would integrate 13^x-11^x.
lets say i integrate 13^x, would that be 13^x*log 13. i know there is a formula for that but i don't remember
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