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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The general solution of the differntial equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{coses}\ \text{x}$ is:
  • A
    $\text{x}+\text{y}\sin\text{x}=\text{C}$
  • B
    $\text{x}+\text{y}\cos\text{x}=\text{C}$
  • C
    $\text{y}+\text{x}(\sin\text{x}+\cos\text{x})=\text{C}$
  • ✓
    $\text{y}\sin\text{x}=\text{x}+\text{C}$
✓

Answer

Correct option: D.
$\text{y}\sin\text{x}=\text{x}+\text{C}$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{coses}\ \text{x}$
Comparting with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ we get
$\text{P}=\cot\text{x}$
$\text{Q}=\text{coses} \ \text{x}$
Now,
$\text{I.F}=\text{e}^{\int\cot\text{x}\text{dx}}$
$=\text{e}^{\log(\sin\text{x})}$
$=\sin\text{x}$
So, the solution is given by
$\Rightarrow \text{y}\sin\text{x}=\int\sin\text{x}\times\text{cosec}\ \text{x}\text{dx}+\text{C}$
$\Rightarrow \text{y}\sin\text{x}=\text{x}+\text{C}$

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