find the equation of the curve which passes through the pts (2,-2) and (4,2) and for which dy/dx =x^2(x-k) where k is a constant.

Question

anonymous · Answer

is there a graph with this question?

anonymous · Answer

no

anonymous · Answer

ok

anonymous · Answer

i actually don't know this but i will see if any of my friends know this and help

amistre64 · Answer

integrate, and use the intial values to define the curve

amistre64 · Answer

dy/dx =x^2(x-k)

dy =x^2(x-k) dx

$$\int~dy =\int~x^2(x-k) dx$$

anonymous · Answer

i did 
i cant the right answer.

anonymous · Answer

what about the k?

amistre64 · Answer

what was your integration result?

amistre64 · Answer

the k is some constant, you are given 2 points to initialize so that it forms a system of 2 equations in 2 unknowns

amistre64 · Answer

let k=3 for the moment if its distracting you

anonymous · Answer

oh ok

anonymous · Answer

when i tried to integrate with the k ,i got
1/4(x^4) -1/3 k (x^3)

amistre64 · Answer

$$\int~dy =\int~x^2(x-k) dx$$
$$y =\int~x^3-kx^2~dx$$
$$y =\frac14x^4-\frac13kx^3+C$$
initialize it with the points  (2,-2) and (4,2) 

$$-2 =\frac142^4-\frac k32^3+C$$
$$2 =\frac144^4-\frac k34^3+C$$
solve for k and C

anonymous · Answer

ok i get it know

amistre64 · Answer

by elimination,  -1 the top and add

$$2 =-\frac142^4+\frac k32^3-C$$$$2 =\frac144^4-\frac k34^3+C$$
$$4=\frac14(4^4-2^4)+k(\frac13(2^3-4^3))$$
$$4-\frac14(4^4-2^4)=k(\frac13(2^3-4^3))$$
$$\frac{4-\frac14(4^4-2^4)}{\frac13(2^3-4^3)}=k$$

anonymous · Answer

k=3
and then i replace in the other eqn to find C....

anonymous · Answer

tysm i got the answer right :)

amistre64 · Answer

:)  youre welcome