Q.$\left(1-x^2\right)\dfrac{dy}{dx}-xy=1$

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. $\ln|y|=-\int p(x)\,dx+k$.
4. Exponentiating: $y=e^{k}\,e^{-\int p(x)\,dx}$.
5. Writing $c=e^{k}$ as an arbitrary constant, $y=ce^{-\int p\,dx}$.

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: $\int P\,dx=\ln(\sin x)$.
4. Differentiating both sides with respect to $x$: $P=\dfrac{d}{dx}\big(\ln\sin x\big)=\dfrac{\cos x}{\sin x}$.
5. This simplifies to $P=\cot x$.

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: $\int P\,dx = \ln(\cos x)$.
4. Differentiate both sides with respect to $x$ to recover $P$: $P = \dfrac{d}{dx}\big[\ln(\cos x)\big] = \dfrac{1}{\cos x}\cdot(-\sin x) = -\tan x$.

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.
5. (The given condition $m<0$ tells us the exponent $-my$ actually grows with $y$, but it does not change the functional form of the solution.)

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 $\text{I.F.} = e^{\int P\,dx} = e^{\int(-\tan x)\,dx} = e^{-\int\tan x\,dx}$.
4. Using the standard integral $\int\tan x\,dx = \ln|\sec x| = -\ln|\cos x|$: $\text{I.F.} = e^{-(-\ln|\cos x|)} = e^{\ln|\cos x|} = \cos x$
5. So the integrating factor is $\cos x$, matching option (b).