Q.$(x+3y^2)\frac{dy}{dx}=y \quad (y>0)$

The equation is linear in $x$ when we treat $x$ as a function of $y$ (given $y>0$).

Flip the derivative: $(x+3y^2)\frac{dy}{dx}=y \implies \frac{dx}{dy} = \frac{x+3y^2}{y} = \frac{x}{y} + 3y$, i.e. $\frac{dx}{dy} - \frac{1}{y}x = 3y$.

Integrating factor: $\mu = e^{-\int\frac{1}{y}dy} = e^{-\log y} = \frac{1}{y}$.

Integrate: $\frac{d}{dy}\left(\frac{x}{y}\right) = 3 \implies \frac{x}{y} = 3y + C$. …

Reading the problem

The trailing sentence (“find a particular solution satisfying the given condition”) is a heading carried over from a set of exercises, but no actual condition is attached to this equation. The honest result is therefore the general solution; if a condition such as $x(y_0)=x_0$ were given, we would substitute it at the end to fix $C$.

Spotting the trick

$(x+3y^2)\frac{dy}{dx}=y$ is not linear in $y$, but flipping the derivative makes it linear in $x$ (valid since $y>0$).

Set up

Integrating factor …

Method: Linear-in-$x$ equation, then apply a condition if given

When $\dfrac{dy}{dx}$ makes the equation non-linear but flipping to $\dfrac{dx}{dy}$ is linear, solve it that way and only afterwards use any initial condition.

Steps

Step 1: Flip the derivative.

From the given equation obtain $\dfrac{dx}{dy}+P(y)\,x=Q(y)$.

Step 2: Integrating factor in $y$.

For $P=-\dfrac{1}{y}$ this is $\dfrac{1}{y}$.

Step 3: Integrate. …

Common Mistakes

Mistake 1: Trying to treat it as linear in $y$.

Why it's wrong: $(x+3y^2)\dfrac{dy}{dx}=y$ is non-linear in $y$; flipping to $\dfrac{dx}{dy}-\tfrac1y x=3y$ makes it linear in $x$. Correct approach: use $x(y)$ (valid since $y>0$).

Mistake 2: Integrating-factor sign from $P=-\tfrac1y$.

Why it's wrong: $\int-\tfrac1y\,dy=-\log y$, so I.F. $=\tfrac1y$, not $y$. Correct approach: exponentiate $-\log y$. …