Q.Find the general solution of $(1+\tan y)(dx-dy)+2x\,dy=0$.

We treat this as an Initial Value Problem in $x(y)$ — the equation mixes $dx$ and $dy$, so we rearrange to get a linear first-order ODE in $x$.

Step 1: Expand and group terms

Step 2: Write as $\frac{dx}{dy}$

Divide through by $1+\tan y$:

Step 3: Solve the linear ODE

Integrating factor:

Use $\int \frac{dy}{1+\tan y} = \frac{y}{2} + \frac12 \log|\cos y + \sin y|$ (standard result).

Thus $\mu(y) = e^{y + \log|\cos y + \sin y|} = e^y (\cos y + \sin y)$.

Multiply through:

Integrate: …

Put in linear form. Divide $(1+\tan y)(dx-dy)+2x\,dy=0$ by $dy$:

Integrating factor. Since $\dfrac{2}{1+\tan y} = \dfrac{2\cos y}{\cos y+\sin y} = 1 + \dfrac{\cos y - \sin y}{\cos y+\sin y}$,

Multiply and integrate. The left side is an exact derivative:

Since $\dfrac{d}{dy}\big(e^{y}\sin y\big) = e^{y}(\sin y + \cos y)$,

Solve for $x$: …

Method: Rearranging a mixed-differential equation to linear in $x$

Use this when both $dx$ and $dy$ appear with a trig coefficient, and the equation becomes linear once $x$ is taken as dependent.

Steps

Step 1: Group $dx$ and $dy$, then form $\dfrac{dx}{dy}$

Expand the products, collect terms, and write

Step 2: Integrating factor in $y$

$\mathrm{IF}=e^{\int P(y)\,dy}.$ …

Common Mistakes

Mistake 1: Not regrouping the mixed differentials

Why it's wrong: $(1+\tan y)(dx-dy)+2x\,dy=0$ must be collected into $(1+\tan y)\frac{dx}{dy}+2x=1+\tan y$ before solving. Correct approach: group $dx$ and $dy$ terms.

Mistake 2: Treating it as linear in $y$

Why it's wrong: it is linear in $x$; use $\frac{dx}{dy}+\frac{2}{1+\tan y}x=1$. Correct approach: take $x$ as dependent. …