Question

Solve the initial-value problem (4x^3*y^2-6x^2*y-2x-3)dx+(2x^4*y-2x^3)dy=0, y(1)=3.

Answer 1

I've got the answer, which is x^4*y^2-2x^3*y-x^2-3x=C but the initial-value is y(1)=3, how do I solve for y?

Answer 2

@hartnn @myininaya

Answer 3

y(1) =3
means
when x=1, y=3

Answer 4

So I found that C=-1 after I plug x=1 and y=3. But how do I solve for y?

Answer 5

why do u need y ?

Answer 6

its not gonna be easy, if possible

Answer 7

Because the answer in the book is in the form of y=...

Answer 8

The answer in the book is...\[y=\frac{ x+\sqrt{2x ^{2}+3x-1} }{ x ^{2} }\]

Answer 9

a =x^4
b=-2x^3
c= -x^2-3x+1

then you have
ay^2+by+c =0

Answer 10

quadratic formula

Answer 11

you sure your answer is correct ?

Answer 12

Well, I guess I'm pretty sure. Let me work the quadratic formula out.

Answer 13

i sense you'll get the correct answer :)

Answer 14

Thank you so much! I got the right answer!