Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$
$\frac{\text{dy}}{\text{dx}}-\text{y}\tan\text{x}=-2\sin\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=-\tan\text{x}$
$\text{Q}=-2\sin\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\tan\text{xdx}}$
$=\text{e}^{-\log|\sec\text{x}|}=\cos\text{x}$
Multiplying both sides of (1) by $\cos\text{x},$ we get
$\cos\text{x}\Big(\frac{\text{dy}}{\text{dx}}-\text{y}\tan\text{x}\Big)=-2\sin\text{x}\times\cos\text{x}$
$\Rightarrow\ \cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=-\sin2\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\cos\text{x}=-\int\sin2\text{x dx + C}$
$\Rightarrow\ \text{y}\cos\text{x}=\frac{\cos2\text{x}}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=\frac{1-2\sin^2\text{x}}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=-\sin^2\text{x}+\frac{1}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=-\sin^2\text{x}+\text{K}$ $\Big($where $\text{K}=\frac{1}2+\text{c}\Big)$
$\Rightarrow\ \text{y}=\sec\text{x}\big(-\sin^2\text{x}+\text{K}\big)$
Hence, $\text{y}=\sec\text{x}\big(-\sin^2\text{x}+\text{K}\big)$ is the required solution.
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$
$\frac{\text{dy}}{\text{dx}}-\text{y}\tan\text{x}=-2\sin\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=-\tan\text{x}$
$\text{Q}=-2\sin\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\tan\text{xdx}}$
$=\text{e}^{-\log|\sec\text{x}|}=\cos\text{x}$
Multiplying both sides of (1) by $\cos\text{x},$ we get
$\cos\text{x}\Big(\frac{\text{dy}}{\text{dx}}-\text{y}\tan\text{x}\Big)=-2\sin\text{x}\times\cos\text{x}$
$\Rightarrow\ \cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=-\sin2\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\cos\text{x}=-\int\sin2\text{x dx + C}$
$\Rightarrow\ \text{y}\cos\text{x}=\frac{\cos2\text{x}}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=\frac{1-2\sin^2\text{x}}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=-\sin^2\text{x}+\frac{1}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=-\sin^2\text{x}+\text{K}$ $\Big($where $\text{K}=\frac{1}2+\text{c}\Big)$
$\Rightarrow\ \text{y}=\sec\text{x}\big(-\sin^2\text{x}+\text{K}\big)$
Hence, $\text{y}=\sec\text{x}\big(-\sin^2\text{x}+\text{K}\big)$ is the required solution.
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