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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS3 Marks
Question
For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}+\text{y}=1(\text{y}\neq1)$
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Answer

The given differential equation is
$\frac{\text{dy}}{\text{dx}}+\text{y}=1\ \text{or} \ \frac{\text{dy}}{\text{dx}}=1-\text{y}$
Separating the variables, we get,
$\frac{1}{1-\text{y}}\text{dy}=\text{dx}$
Integrating, $\int\frac{1}{1-\text{y}}\text{dy}=\int1\ \text{dx}$
$\therefore \ \frac{\text{log|1-y|}}{-1}= \text{x+c}'$
$\therefore \ \text{log|1-y|}=-\text{x}-\text{c}'$
$\therefore\ |1-\text{y}|=\text{e}^{-\text{x}-\text{c}'}\ \text{or}\ |1-\text{y}|=\text{e}^{-\text{x}}.\text{e}^{-\text{c}}$
$\therefore \ 1-\text{y}=\pm\text{e}^{-\text{e}'}\ \text{e}^{-\text{x}}$
$\therefore \ 1-\text{y}=-\text{ce}^{-\text{x}} \ \ \text{where}-\text{c}=\pm\text{e}^{-\text{x}}$
$\therefore\ \text{y}=1+\text{ce}^{-\text{x}}$ is the required solution.

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