Question

dy/dx = x general solution

Answer 1

dy/dx = x

dy = x*dx

integrate both sides with respect to the variables on each side to get

int(dy) = int(x*dx)

y = (1/2)x^2 + C

Answer 2

notice how if you derive both sides with respect to x, you will get

y = (1/2)x^2 + C

d/dx[y] = d/dx[ (1/2)x^2 + C ]

d/dx[y] = d/dx[ (1/2)x^2 ] + d/dx [ C ]

dy/dx = 2*(1/2)x^(2-1) + 0

dy/dx = (2/2)x^(1) + 0

dy/dx = x

Answer 3

so thats the general solution, so if it gives me the point (1,2) i just plug in x and y to find c?

Answer 4

yes exactly

Answer 5

the general solution is basically a collection of functions (with similar properties)

a specific or particular solution is a single function that goes through that point

Answer 6

so the particluar solution is y= .5x^2 +(3/2)

Answer 7

correct since that function has the point (1,2) on it

Answer 8

okay what about dy/dx = 6x-xy for the general solution?

Answer 9

hmm this doesn't look separable

Answer 10

if it were separable, it would be in the form

dy/dx = f(x) * g(y)

Answer 11

would i take out an x and then divide both sides by 6-y

Answer 12

or how about this one. seems easier dy/dx = xy

Answer 13

you would first separate out the x and y variables

Answer 14

dy/dx = xy

dy/y = x*dx

Answer 15

then integrate both sides with respect to the variables on each side to get

Answer 16

so the general solution is lnIyI = .5x^2+c

Answer 17

which turns into

|y| = e^(0.5x^2 + C)

y = +-e^(0.5x^2 + C)

you can rearrange terms a bit, but that's the basic general solution

Answer 18

if it gives me f(0)=.5 when i plug in i get lnI.5I= c so what is ln of .5

Answer 19

dy/dx = x/y^2

Answer 20

lnIyI = .5x^2+c

lnI0.5I = .5(0)^2+c

c = ln|0.5|

c = -0.693147

Answer 21

dy/dx = x/y^2

y^2*dy =x*dx

Answer 22

did you just type that into the calculator?

Answer 23

yes

Answer 24

i used a calc to find ln|0.5| = -0.693147

Answer 25

so how would in integrate y^2dy=xdx

Answer 26

integral of y^2*dy is ???

Answer 27

use the rule

integral of y^n*dy = ( 1/(n+1) ) *y^(n+1) + C

Answer 28

dw/dt=1/5(w-200)

Answer 29

general solution

Answer 30

w-200 is in the denominator? or no?

Answer 31

no its part of the numerator

Answer 32

ok just checking

Answer 33

its multiplied by 1/5

Answer 34

dw/dt=1/5(w-200)

dw/(w-200)=dt/5

integrate both sides

Answer 35

use u-substitution to integrate the left side

Answer 36

how would i integrate the right side

Answer 37

dt/5 is the same as (1/5)*dt

Answer 38

so you can pull the coefficient 1/5 out and take the integral of dt

Answer 39

or 1dt

Answer 40

so what would that be?

Answer 41

int((1/5)*dt)

(1/5)*int(dt)

(1/5)*(t + C)

(1/5)t + C

Answer 42

so the general solution would be lnIw-200I =(1/5)t+c

Answer 43

yep, you could optionally solve for w, but you pretty much got it