Question

solve the initial value problem dy/dx=2yx-x^2y and determine where the solution attains its minimal value

Answer 1

dy = (2yx - x^2y) dx
dy = y(2x - x^2) dx
dy / y = (2x - x^2) dx
ln(y) = x^2 - x^3 / 3
y = e^((x^2 - x^3) / 3) + C

Answer 2

but how to get the value of c

Answer 3

C is a constant - it can be anything.

Answer 4

Because if you differentiate a constant, you get 0

Answer 5

the value of "C" will depend on the point that is provided in the initial question. Otherwise, "C" is any constant....

Answer 6

but i got to solve it by separable equation

Answer 7

yes y(0)=3

Answer 8

Right, right - plug in that initial to find C.

Answer 9

i will type the question again

Answer 10

there are an infinite number of "graphs" that have the same derivitive. They all differ by the value of their constant (C).

the only way to determine the value of (C) is to have information regarding some point onthe graph that contains (C) to anchor it in...

Answer 11

I saw the question. You need your initial value (x = 0 y = __)

Answer 12

solve the initial value problem dy/dx=2y^2+xy^2, y(0)=1 and determine where the solution attain its minimal value
i am sending the same example and solution i got to do the same with the above one

Answer 13

im sending the solution now

Answer 14

solution;
separating variables gives; y^-2dy=(2+x)dx

Answer 15

Get out of here if you're not willing to listen to us.

Answer 16

-y^-1=2x+x^2/2+C

Answer 17

OK GO ON

Answer 18

so explain me plz