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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
Question
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:

  1. $\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$

  2. $\text{x}(\text{y}-\cos\text{x})=\sin\text{x}+\text{c}$

  3. $\text{x}\text{y}\cos\text{x}=\sin\text{x}+\text{c}$

  4. $\text{x}(\text{y}+\cos\text{x})=\cos\text{x}+\text{c}$

✓

Answer

  1. $\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$

Solution:

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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