Question

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=\frac{4 y^3+2 y x^2}{3 x y^2+x^3}, y(1)=1$ If for some $n \in N , y (2) \in[ n -1, n )$, then $n$ is equal to ______

Answer 1

The correct answer is 3
dxdy​=3xy2+x34y3+2yx2​,y(1)=1
dxdy​=3(y/x)2+14(y/x)3+2(y/x)​
y=xp
xdxdp​+p=3p2+14p3+2p​
xdxdp​=3p2+1p3+p​
∫p3+p3p2+1​dp=∫xdx​
ln(p3+p)=lnx+lnC
p3+p=xC
(xy​)3+(xy​)=xC
y3+x2y=x4C
x=1,y=1
1+1=C⇒C=2
y3+x2y=2x4
Put x=2
y3+4y−32=0
Having root between 2 and 3
y(2)∈[2,3)