Question

given dy/dx=(3x^2+4x)/y. if the point (1, sqrt(10)) is on the graph relating x and y, then what is y when x=0?
I read through a similar answer but I can't understand the work at the beginning at all. My teacher never taught us this type of problem and put this on the test. I'll understand it best if I can see all the work at once and then ask questions. :) Thanks so much to anyone who helps out.

Answer 1

What level are you studying? highschool, uni?

Answer 2

High school, but it's a college-level class. AP calculus.

Answer 3

ok

Answer 4

This is a diferential equation... It gives you the slope of the graph relating x and y. To solve it you have to find this graph. To do so, put all that depends on y to the left side and what depends on x to the right one and integrate. You will get a equation and a undefind constant. To figure out the value of the constant use the information about that point (1, sqrt(10)) should be on this graph. If have more quastions tell me and i solve it for you

Answer 5

ydy=(3x^2+4x)dx
Integrating:
1/2y^2 = x^3 +2x^2 +C
y= 2sqrt(x^3 +2x^2 +C)
Now to find the constant substitute values of x,y for (1,sqrt(10)) and solve for C
sqrt(10)=2sqrt(1+2+C)
C=(10-6)/2 = 2
So you equation relating x and y is
y= 2sqrt(x^3 +2x^2 +2)
The value of y when x=0 is y=2sqrt(2)

Answer 6

check calculations just in case....:)

Answer 7

I'm just so lost on this problem right now :( .. How would I go about separating what's dependent on either variable? the only time she's had us do that before were in very much simpler cases like 3y^3+2x^2+24xy =13 or something.

Answer 8

Hmm. You know integration?

Answer 9

Becouse onestly, I am trying to think other way to solve it, but can't find any