Q.$\left(1+x+xy^2\right)\dfrac{dy}{dx}+\left(y+y^3\right)=0$

Separating variables in $\dfrac{dy}{dx}+p(x)y=0$ and integrating gives $y=ce^{-\int p\,dx}$.

1. Start with $\dfrac{dy}{dx}+p(x)y=0$, i.e. $\dfrac{dy}{dx}=-p(x)y$.
2. Separating variables: $\dfrac{dy}{y}=-p(x)\,dx$ (assuming $y\ne0$).
3. Integrating both sides: $\displaystyle\int\dfrac{dy}{y}=-\int p(x)\,dx+k$, i.e. $\ln|y|=-\int p(x)\,dx+k$. …

Since the integrating factor $e^{\int P\,dx}$ equals $\sin x$, differentiating $\ln\sin x$ gives $P=\cot x$.

1. For the linear equation $\dfrac{dy}{dx}+Py=Q$, the integrating factor is $\text{I.F.}=e^{\int P\,dx}$.
2. We are told $\text{I.F.}=\sin x$, so $e^{\int P\,dx}=\sin x$.
3. Taking natural logarithms of both sides: $\int P\,dx=\ln(\sin x)$. …

Since the integrating factor is $e^{\int P\,dx}$ and this is given as $\cos x$, differentiating $\ln(\cos x)$ recovers $P=-\tan x$.

1. For a linear first-order equation $\dfrac{dy}{dx}+Py=Q$, the integrating factor (I.F.) is $\mathrm{I.F.} = e^{\int P\,dx}$.
2. We are told $\mathrm{I.F.} = \cos x$, so $e^{\int P\,dx} = \cos x$.
3. Taking natural logs of both sides: $\int P\,dx = \ln(\cos x)$. …

Separating variables in $\dfrac{dx}{dy}+mx=0$ gives the general solution $x=ce^{-my}$.

1. Rewrite the equation: $\dfrac{dx}{dy}=-mx$.
2. Separate variables: $\dfrac{dx}{x}=-m\,dy$.
3. Integrate both sides: $\ln x = -my + k$, where $k$ is a constant of integration.
4. Exponentiate: $x = e^{-my+k}=ce^{-my}$, where $c=e^k$ is an arbitrary constant. …

Identify the equation as first-order linear $\frac{dy}{dx}+Py=Q$ with $P=-\tan x$, then compute $\text{I.F.}=e^{\int P\,dx}$, using $\int\tan x\,dx=\ln|\sec x|$ and simplifying $e^{-\ln|\sec x|}=\cos x$.

1. Given: $\dfrac{dy}{dx} - y\tan x = \cos x$.
2. This is in the standard linear form $\dfrac{dy}{dx} + Py = Q$, with $P = -\tan x$ and $Q=\cos x$.
3. The integrating factor is $\text{I.F.} = e^{\int P\,dx} = e^{\int(-\tan x)\,dx} = e^{-\int\tan x\,dx}$. …