Question

If the curve satisfying $xdx=\left(\frac{x^2}{y}-y^3\right)dy$ passes through $(0,2)$ then the value of $\left(y\right(4{)}^2(4-(y\left(4\right){)}^2)$

Answer 1

Given : $xdx=\left(\frac{x^2}{y}-y^3\right)dy$

$⟹(\frac{x^2}{y}-y^3)\frac{dy}{dx}=x$ ........... $\left(1\right)$
Let $R(x,y)=-x$ and $S(x,y)=\frac{x^2}{y}-y^3$
This is not an exact equation, because
$\frac{d\left(R(x,y)\right)}{dx}=-1$ and $\frac{2x}{y}=\frac{d\left(S(x,y)\right)}{dx}$
Therefore taking integrating factor $u$ above become exact.
$\Rightarrow \frac{d\left(uR(x,y)\right)}{dx}=\frac{d\left(uS(x,y)\right)}{dx}\Rightarrow \frac{\frac{du}{dx}}{u}=-\frac{2}{y}$
Integrating both sides w.r.t x
$\log u=-2\log y⟹\log u=-\log y^2⟹u=\frac{1}{y^2}$
Multiplying both sides of $\left(1\right)$ by $u$, we get
$\frac{1}{y^2}(\frac{x^2}{y}-y^3)\frac{dy}{dx}=\frac{x}{y^2}$
Let $P(x,y)=-\frac{x}{y^2}$ and $Q(x,y)=\frac{x^2}{y}-y^3$
Define $f(x,y)$ such that $\frac{d\left(f(x,y)\right)}{dx}=P(x,y)$ and $\frac{d\left(f(x,y)\right)}{dy}=Q(x,y)$
Then, the solution will be given by $f(x,y)=c$ where $c$ is constant
Now,
Integrating $\frac{d\left(f(x,y)\right)}{dx}$ w.r.t $x$ in order to find $f(x,y)$
$f(x,y)=-\frac{x^2}{2y^2}+g\left(y\right)$
where $g\left(y\right)$ is an arbitrary function of y.
Differentiating $f(x,y)$ w.r.t to $y$ in order to find $g\left(y\right)$
$\frac{d\left(f(x,y)\right)}{dx}=\frac{d}{dy}(-\frac{x^2}{2y^2}+g\left(y\right))=\frac{x^2}{y^3}+\frac{dg\left(y\right)}{dy}$
Substituting $\frac{d\left(f(x,y)\right)}{dx}$ in $Q(x,y)$
$\frac{x^2}{y^3}+\frac{dg\left(y\right)}{dy}=-y+\frac{x^2}{y^3}\Rightarrow \frac{dg\left(y\right)}{dy}=-y$
Integrate $\frac{dg\left(y\right)}{dy}$ w.r.t y
$g\left(y\right)=-\frac{y^2}{2}$
Substitute $g\left(y\right)$ in $f(x,y)$
$f(x,y)=-\frac{y^2}{2}-\frac{x^2}{2y^2}$
Then solution is
$-\frac{y^2}{2}-\frac{x^2}{2y^2}=c$
It passes through $(0,2)$ gives $c=-2$
Then for $x=4$
${\left(y\left(4\right)\right)}^2(4-{\left(y\left(4\right)\right)}^2)=16$