Solution of the differential equation dy dx + y x = is:

MCQ

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
$\text{x}(\text{y}+\cos\text{x})=\cos\text{x}+\text{C}$
D
None of these.

✓

Answer

Correct option: A.

$\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$

We have,
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\sin\text{x}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}=\sin\text{x}\ ...(\text{i})$
Comparing with we get,
$\text{P}=\frac{1}{\text{x}}$
$\text{Q}=\sin\text{x}$
Now,
$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log|\text{x}|}$
$=\text{x}$
Therefore, intergration of $(i)$ is given by,
$\text{y}\times\text{I.F}=\int\text{x}^{2}\times\text{I.F.}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}=\int\text{x}\ \sin\text{x}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}=\text{x}\int\sin\text{x}\ \text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x}\ \text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\text{yx}=-\text{x}\cos\text{x}+\int\cos\text{x}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}+\text{x}\cos\text{x}=\sin\text{x}+\text{C}$
$\Rightarrow\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$

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