Q.Solve the differential equation $\dfrac{dy}{dx} + 3y = e^{2x}$.

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 $P=1,\ Q=e^{-y}\sec^2y$.

Integrating factor: $e^{\int1\,dy}=e^y$.

Multiply through by the integrating factor and integrate:

$x\cdot e^y = \int e^y\cdot e^{-y}\sec^2y\,dy = \int\sec^2y\,dy = \tan y + C$

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:

$x\cdot\frac1y = \int 2y^2\cdot\frac1y\,dy = \int 2y\,dy = y^2+C$

So $\dfrac{x}{y} = y^2+C$, i.e. $x = y^3+Cy$.

Apply the initial condition $x=1$ when $y=-1$:

$1 = (-1)^3 + C(-1) = -1-C \ \Rightarrow\ C = -2$

So the particular solution is:

$x = y^3-2y$