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