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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS3 Marks
Question
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+\sec\text{xy}=\tan\text{x}\Big(0\leq\text{x}<\frac{\pi}{2}\Big)$
✓

Answer

The given differential equation is:$\frac{\text{dy}}{\text{dx}}+​​\text{py}=\text{Q (where p}=\sec\text{x and Q}=\tan\text{x})$
$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int\sec\text{x dx}}=\text{e}^{\log(\sec\text{x}+\tan\text{x})}=\sec.\text{x}+\tan\text{x}.$
The general solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\int\tan\text{x}(\sec\text{x}+\tan\text{x})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\int\sec\text{x}\tan\text{x}\text{dx}+\int\tan^2\text{x}\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\sec\text{x}+\int(\sec^2 \text{x}-1)\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\sec\text{x}+\tan\text{x}-\text{x}+\text{C}$
This is the required general solution of the given differential equation.

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