Solve (2x^3y+1) dx + x^4 dy + x^2tan z dz = 0
Solve \[ (2x^3y+1)\,dx + x^4\,dy + x^2\tan z\,dz = 0. \]
Explanation of Method
Check exactness of $P\,dx+Q\,dy+R\,dz=0$; since it fails, look for an integrating factor depending on a single variable, restore exactness, and integrate the resulting exact equation to a potential $F(x,y,z)=C$.
Step 1 · Check exactness
With $P=2x^3y+1$, $Q=x^4$, $R=x^2\tan z$: \[ P_y=2x^3,\qquad Q_x=4x^3. \] Since $P_y\neq Q_x$, the equation is not exact.
Step 2 · Find an integrating factor $\mu(x)$
Impose $(\mu P)_y=(\mu Q)_x$: \[ 2\mu x^3 = \mu' x^4+4\mu x^3 \implies \frac{\mu'}{\mu}=-\frac{2}{x} \implies \mu = x^{-2}. \]
Step 3 · Multiply through and verify exactness
\[ \left(2xy+\frac{1}{x^2}\right)dx + x^2\,dy + \tan z\,dz = 0. \] With $\tilde P=2xy+x^{-2}$, $\tilde Q=x^2$, $\tilde R=\tan z$: \[ \tilde P_y=2x=\tilde Q_x,\qquad \tilde P_z=0=\tilde R_x,\qquad \tilde Q_z=0=\tilde R_y. \checkmark \]
Step 4 · Find the potential $F$
From $F_y=x^2$: $F=x^2y+g(x,z)$. From $F_x=2xy+x^{-2}$: $g_x=x^{-2}$, so $g=-\dfrac{1}{x}+h(z)$. Hence \[ F = x^2y-\frac{1}{x}+h(z). \] From $F_z=\tan z$: $h'(z)=\tan z$, so $h(z)=-\ln|\cos z|+C_0$.
Verification
With $F=x^2y-\dfrac1x-\ln|\cos z|$: \[ F_x = 2xy+\frac{1}{x^2},\qquad F_y=x^2,\qquad F_z=\tan z. \] Multiplying $dF=0$ by $x^2$ recovers exactly the original equation: \[ (2x^3y+1)\,dx+x^4\,dy+x^2\tan z\,dz=0. \checkmark \]
Result
The general integral is: