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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 a particular solution satisfying the given condition:
$\frac{\text{dy}}{\text{dx}}=\text{y}$ tan x; y = 1 when x = 0
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Answer

The given differential equation is
$\frac{\text{dy}}{\text{dx}}=\text{y tan x}$
Separating the variables, we get,

$\frac{1}{\text{y}}\text{dy}= \text{tanx}\ \text{dx}$
$\text{Integrating,}\ \int\frac{1}{\text{y}}\text{dy}=\int\text{tanx dx}$
$\therefore\ \log|\text{y}|=-\log|\cos{\text{x}}|+\text{c'}\ \ ... (1)$
$\therefore\ \log|\text{y}|+\log|\cos{\text{x}}|=\text{c'}$
$\therefore\ \log|\text{y}\cos{\text{x}}|=\text{c'}$
$\therefore|\text{y}\cos{\text{x}}|=\text{e}^{\text{c'}}$
$\therefore\ \text{y}\cos{\text{x}}=\pm\text{e}^{\text{c}}\ ​​\text{or}\ \text{y cosx}=\text{c}$
Now y = 1 when x = 0
$\therefore1\cos0=\text{c}\ \Rightarrow\ 1\times1=\text{c}\Rightarrow\text{c}=1$
$\therefore\text{form (1),}\ \text{y}\cos\text{x}=1,$ which is required solution.

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