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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
Choose the correct answer from the given four options.Solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\sin\text{x}$ is:
  • ✓
    $\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$
  • B
    $\text{x}(\text{y}-\cos\text{x})=\sin\text{x}+\text{c}$
  • C
    $\text{x}\text{y}\cos\text{x}=\sin\text{x}+\text{c}$
  • D
    $\text{x}(\text{y}+\cos\text{x})=\cos\text{x}+\text{c}$
✓

Answer

Correct option: A.
$\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$
Given differential equation is
$\frac{\text{dy}}{\text{dx}}+\text{y}\frac{1}{\text{x}}=\sin\text{x}$
Which is liner differential equations.
Here, $\text{P}=\frac{1}{\text{x}}$ and $\text{Q}=\sin\text{x}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log\text{x}}=\text{x}$
The general solution is
$\text{yx}=\int\text{x}\sin\text{xdx}+\text{c}\ .....(\text{i})$
Take $\text{I}=\int\text{x}\sin\text{xdx}$
$-\text{x}\cos\text{x}-\int-\cos\text{xdx}$
$=-\text{x}\cos\text{x}+\sin\text{x}$
Put the value of $1$ in $Eq. (i),$ we get
$\text{xy}=-\text{x}\cos\text{x}+\sin\text{x}+\text{c}$
$\Rightarrow\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$

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