Q.Solve the differential equation $\dfrac{dy}{dx} + y\tan x = \cos x$, $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$.

Treat $x$ as the dependent variable (function of $y$) to get a linear first-order equation.

Given $e^{-y}\sec^2y\,dy = dx+x\,dy$. Rearranging:

$e^{-y}\sec^2y\,dy - x\,dy = dx \;\Rightarrow\; \frac{dx}{dy} = e^{-y}\sec^2y - x \;\Rightarrow\; \frac{dx}{dy}+x = e^{-y}\sec^2y$

This is a linear differential equation in $x$ (as a function of $y$), of the form $\dfrac{dx}{dy}+Px=Q$ with $P=1,\ Q=e^{-y}\sec^2y$.

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Flip the equation to treat $x$ as a function of $y$; it becomes linear in $x$, solvable with an integrating factor.

Given $(x+2y^3)\dfrac{dy}{dx} = y$, given $x=1$ when $y=-1$.

Rewrite as $\dfrac{dx}{dy} = \dfrac{x+2y^3}{y} = \dfrac{x}{y}+2y^2$, i.e.:

$\frac{dx}{dy} - \frac{x}{y} = 2y^2$

This is a linear first-order ODE in $x$ (as a function of $y$), of the form $\dfrac{dx}{dy}+Px=Q$ with $P=-1/y$, $Q=2y^2$.

Integrating factor: $\displaystyle \text{I.F.} = e^{\int -1/y\,dy} = e^{-\ln y} = \frac1y$.

Multiplying through and integrating: …