How do I solve $(3x^2 \tan y- 2y^3/x^3 )dx + (4y^3 + x^3 \cos^2 y+ 3y^2/x^2 ) dy=0$ Somebody please help me to solve this Differential equation.
$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0 $
 A: It is an exact DE, solution is $x^3\tan{y}+y^4+\frac{y^3}{x^2}+C=0$
A: Claim:  This is an exact first order ordinary differential equation.
To verify this claim, we impose some standard notation, then perform a simple test.  Let
\begin{align*}
M(x,y) &= 3x^2 \tan(y) - \frac{2y^3}{x^3}  \text{,}  \\
N(x,y) &= 4 y^3 + \frac{x^3}{\cos^2 y} + \frac{3y^2}{x^2}
\end{align*}
(Notice that, for a while, we will not treat $y$ as a dependent variable.)
We wish to find an $f(x,y)$ such that both
\begin{align*}
\frac{\partial}{\partial x} f(x,y) &= M(x,y)  \\
\frac{\partial}{\partial y} f(x,y) &= N(x,y)
\end{align*}
If so, the equation is exact.  We perform the test:
$$  \frac{\partial}{\partial y} M(x,y) \overset{?}= \frac{\partial}{\partial x} N(x,y)  $$
(This test verifies $\frac{\partial^2}{\partial y \, \partial x} f(x,y) = \frac{\partial^2}{\partial x \, \partial y} f(x,y)$, which is true under mild assumptions about the various partial derivatives of $f$.)
So we calculate \begin{align*}
\frac{\partial}{\partial y} M(x,y) &= \frac{-6y^2}{x^3} + 3x^2 \sec^2(y)  \text{,}  \\
\frac{\partial}{\partial x} N(x,y) &= \frac{-6y^2}{x^3} + 3x^2 \sec^2(y)  \text{.}
\end{align*}
These are equal, so the equation is exact.
Since the equation is exact, such an $f(x,y)$ exists and the general solution is
$$  f(x,y) + C  \text{.}  $$
We know two partial derivatives of $f$, so we can (partially) recover $f$ by integration.
\begin{align*}
f_1(x,y) &= \int M(x,y) \,\mathrm{d}x  \\
    &= \int 3x^2 \tan(y) - \frac{2y^3}{x^3} \,\mathrm{d}x  \\
    &= \frac{y^3}{x^2} + x^3 \tan(y) + C_1(y)  \text{, and}  \\
f_2(x,y) &= \int N(x,y) \,\mathrm{d}y  \\
    &= \int 4 y^3 + \frac{x^3}{\cos^2 y} + \frac{3y^2}{x^2} \,\mathrm{d}y  \\
    &= \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C_2(x)  \text{.}
\end{align*}
Here, $C_1(y)$ is a function of integration, analogous to the constant of integration.  Notice that if $f(x,y)$ has terms that depend only on $y$ or are constant, those terms are sent to zero in $\frac{\partial}{\partial x} f(x,y)$, so the general antiderivative must represent all possible $f$s differing only by a function of $y$ (and constants).  Similarly, $C_2(x)$ is a function of integration.
Notice that we actually can use $f_2$ to find out about $C_1(y)$.  $C_1(y) = y^4 + C$, where $C$ is a constant of integration.  And reasoning the similarly, $C_2(x) = C$.  (Some describe this process of merging the terms from the two solutions into a single solution that matches both partial derivatives.  Perhaps a better way to describe this is that we are antidifferentiating both sides of the original equation, each with respect to the variable in the differential already present, and then matching the resulting functions of integration.)  So we have
$$  f(x,y) = \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C  \text{.}  $$
Then $f(x,y) = C$ is the general solution, so,
$$  C = \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C  $$
or, what is the same thing (because the range of "an arbitrary number minus another arbitrary number" is an arbitrary number),
$$  C = \frac{y^3}{x^2} + x^3 \tan(y) + y^4  $$
is the general solution to the given equation.

We can, of course, verify that this is the case using implicit differentiation.  First, we stop pretending that $y$ is an independent variable.  Next, we implicitly differentiate the claimed general solution
$$  0 = \frac{x^2 (3y^2 y') - y^3 (2x)}{x^4} + 3x^2 \tan(y) + x^3 \sec^2(y) y' + 4y^3 y'  \text{.}  $$
Then, we solve for $y'$.  \begin{align*}
\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y) &= \frac{x^2 (3y^2 y')}{x^4} + x^3 \sec^2(y) y' + 4y^3 y'  \\
\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y) &= \left( \frac{x^2 (3y^2)}{x^4} + x^3 \sec^2(y) + 4y^3 \right) y'  \\
\frac{\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y)}{\frac{x^2 (3y^2)}{x^4} + x^3 \sec^2(y) + 4y^3} &= y'  \\
\frac{\frac{2 y^3 }{x^3} - 3x^2 \tan(y)}{\frac{3y^2}{x^2} + x^3 \sec^2(y) + 4y^3} &= \frac{\mathrm{d}y}{\mathrm{d}x}  \\
\left(\frac{2 y^3 }{x^3} - 3x^2 \tan(y) \right)\mathrm{d}x &= \left(\frac{3y^2}{x^2} + x^3 \sec^2(y) + 4y^3 \right) \mathrm{d}y  \text{,}
\end{align*}
which should be easy enough to see is the given equation.
A: $$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0$$
$$\left(\tan (y)dx^3+ y^3d\dfrac 1 {x^2}\right)+ \left(dy^4 +{x^3}d{\tan y} + \frac{dy^3}{x^2}\right) =0$$
$$d(\tan (y)x^3)+ d\dfrac {y^3} {x^2}+ dy^4  =0$$
Integrate.
$$x^3\tan (y)+ \dfrac {y^3} {x^2}+ y^4  =C$$
