Question

If the curve satisfying $(x{y}^4+y)dx-xdy=0$ passes through $(1,1)$, then the value of $-41\left(y\right(2){)}^3$ is

• A. 41
• B. 8
• C. 32
• D. 2

Answer 1

Answer: (C)

32

The presence of $ydx-xdy$ terms suggest a factor of the form $\frac{1}{y^{2}f\left(x/y\right)}$.
Dividing both sides by $y^4$, we have
$xdx+\frac{ydx-xdy}{y^4}=0\Rightarrow x^{3}dx+\frac{x^2}{y^2}\frac{ydx-xdy}{y^2}=0\Rightarrow x^{3}dx+{\left(\frac{x}{y}\right)}^{2}d\left(\frac{x}{y}\right)=0$
Integrating
$\frac{x^4}{4}+\frac{1}{3}{\left(\frac{x}{y}\right)}^3=c$
For $y\left(1\right)=1\Rightarrow c=\frac{7}{21}$
Thus
$\frac{8}{{\left(y\left(2\right)\right)}^3}=-\frac{41}{4}\Rightarrow -41{\left(y\left(2\right)\right)}^3=32$