Differential equations — common mistakes

Before finding the integrating factor, always ensure your equation is in the standard form $dy/dx + P(x)y = Q(x)$. If you have, for example, $x \frac{dy}{dx} + 2y = x^2$, you must first divide the entire equation by $x$ to make the coefficient of dy/dx equal to 1.

No. While technically $\int P(x) dx$ has a constant, $e^{P(x)dx + c} = e^{\int P(x)dx} \cdot e^c$. This $e^c$ is just another constant multiplier that would apply to the whole equation and eventually get absorbed into the final constant of integration anyway. So, for simplicity, we omit it when calculating the I.F.

A general solution has the constant +c and represents an entire family of functions that satisfy the differential equation. A particular solution is a single, unique function from that family, found by using a specific point (a boundary condition) to determine the exact value of $c$.

After integrating to get $ln|f(y)| = g(x) + c$, you exponentiate both sides: $|f(y)| = e^{g(x)+c} = e^{g(x)}e^c$. We replace the constant $e^c$ with a new constant, say $A$, so $|f(y)| = Ae^{g(x)}$. This means $f(y) = \pm Ae^{g(x)}$. You can absorb the $\pm$ into the constant $A$. The sign of $A$ is determined by the initial conditions.

This often happens when the initial value is used at different stages. If you solve for the constant of integration $c$ early, you might get $ln(y) = kt + ln(5)$, which gives $y = e^{kt + ln(5)} = e^{kt}e^{ln(5)} = 5e^{kt}$. If you write the general solution as $y = Ae^{kt}$ and then substitute $t=0, y=5$, you get $5 = Ae^0$, so $A=5$. Both methods are correct and lead to the same result.