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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}=\sin\text{x},\text{ y}=0,\text{ when x}=\frac{\pi}{3}$
✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}=\sin\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=2\tan\text{x}$ and $\text{Q}=\sin\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{2\int\tan\text{x dx}}$
$=\text{e}^{2\log|\sec\text{x}|}$
$=\sec^2\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sec^2\text{x},$ we get
$\sec^2\text{x}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}\Big)=\sec^2\text{x}\times\sin\text{x}$
$\sec^2\text{x}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}\Big)=\tan\text{x }\sec\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sec^2\text{x}=\int\tan\text{x}\sec\text{x dx}+\text{C}$
$\text{y}\sec^2\text{x}=\sec{\text{x}}+\text{C}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{3}\Big)=0$
$\therefore\ 0\Big(\sec\frac{\pi}{3}\Big)^2=\sec\frac{\pi}{3}+\text{C}$
$\Rightarrow\text{C}=-2$
Putting the value of C in (2), we get
$\text{y}\sec^2\text{x}=\sec\text{x}-2$
$\Rightarrow\text{y}=\cos\text{x}-2\cos^2\text{x}$
Hence, $\text{y}=\cos\text{x}-2\cos^2\text{x}$ is the required solution.

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