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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:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x }\text{cosec x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x }\text{cosec x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\cot\text{x}$ and $\text{Q}=4\text{x cosec x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cos\text{x dx}}$
$=\text{e}^{\log|\sin\text{x}|}$
$=\sin\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=\sin\text{x}(4\text{x cosec x})$
$\Rightarrow\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=4\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=4\int\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=2\text{x}^2+\text{C}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{2}\Big)=0$
$\therefore\ 0\times\sin\Big(\frac{\pi}{2}\Big)=2\Big(\frac{\pi}{2}\Big)^2+\text{C}$
$\Rightarrow\text{C}=-\frac{\pi^2}{2}$
Putting the value of C in (2) we get
$\text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$
Hence, $\text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$ is the required solution.

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