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.

Question

idealist10 · Answer

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?

idealist10 · Answer

@hartnn @myininaya

hartnn · Answer

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

idealist10 · Answer

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

hartnn · Answer

why do u need y ?

hartnn · Answer

its not gonna be easy, if possible

idealist10 · Answer

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

idealist10 · Answer

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

hartnn · Answer

a =x^4
b=-2x^3
c= -x^2-3x+1

then you have
ay^2+by+c =0

hartnn · Answer

quadratic formula

hartnn · Answer

you sure your answer is correct ?

idealist10 · Answer

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

hartnn · Answer

i sense you'll get the correct answer :)

idealist10 · Answer

Thank you so much! I got the right answer!