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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=2\text{x}+\text{x}^2\tan\text{x},\text{ y}(0)=1$
✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=2\text{x}+\text{x}^2\tan\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q}$
Where $\text{P}=\tan\text{x}$ and $\text{Q}=\text{x}^2\cot\text{x}+2\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\tan\text{x dx}}$
$=\text{e}^{\log|\sec\text{x}|}$
$=\sec\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sec\text{x},$ we get
$\sec\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}\Big)=\sec\text{x}(\text{x}^2\tan\text{x}+2\text{x})$
$\Rightarrow\sec\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}\Big)=\text{x}^2\tan\text{x }\sec\text{x}+2\text{x}\sec\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+2\int\text{x}\sec\text{x dx}\\+2\int\text{x}\sec\text{x dx} +\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+2\sec\text{x}\int\text{x dx}\\-2\int\Big[\frac{\text{d}}{\text{dx}}(\sec\text{x})\int\text{x dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+\text{x}^2\sec\text{x}\\-\int\text{x}^2\tan\text{x }\sec\text{x dx} +\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\text{x}^2\sec\text{x}+\text{C}$
$\Rightarrow\text{y}=\text{x}^2+\text{C}\cos\text{x}\ ...(2)$
Now,
$\text{y}(0)=1$
$\therefore\ 1=0+\text{C}\cos0$
$\Rightarrow\text{C}=1$
Putting the value of C in (2), we get
$\text{y}=\text{x}^2+\cos\text{x}$
Hence, $\text{y}=\text{x}^2+\cos\text{x}$ is the required solution.

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