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

Correct Answer: 3

To solve the given differential equation $\frac{d y}{d x}=\frac{4 y^3+2 y x^2}{3 x y^2+x^3}$ with the initial condition $y(1)=1$, we first attempt separation of variables or find an integrating factor. However, upon inspection, the equation appears complex for straightforward separation. We will instead opt for the method of solving this homogeneous differential equation by substitution.
First, analyze the structure of the equation:

$\frac{d y}{d x}=\frac{4 y^3+2 y x^2}{3 x y^2+x^3}$
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)