Explanation:
(B) : Given,
Differential equation -
\(\left(x-x^{3}\right) d y=\left(y+y x^{2}-3 x^{4}\right) d x, x>2\)
\(\left(x d y-x^{3} d y\right)=y d x+y x^{2} d x-3 x^{4} d x\)
\(x d y-y d x=x^{2}\left(y d x+x d y-3 x^{2} d x\right)\)
\(\frac{x d y-y d x}{x^{2}}=(y d x+x d y)-3 x^{2} d x\)
\(\Rightarrow \quad d\left(\frac{y}{x}\right)=d(x y)-d\left(x^{3}\right)\)
On integrating both sides,
\(\frac{y}{x}=x y-x^{2}+c \tag{i}\)
Given that,
\(y(3)=3\)
\(\frac{3}{3}=3 \times 3-3^{3}+c\)
\(1=9-27+c\)
\(c=19\)
From equation (i),
\(\frac{y}{x}=x y-x^{3}+19\)
Now, At, \(\mathrm{x}=4\)
\(\frac{y}{4}=4 y-(4)^{3}+19\)
\(\frac{y}{4}=4 y-64+19\)
\(y=16 y-180\)
\(15 y=180\)
\(y=12\)
So,
\(y(4)=12\)