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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Find a particular solution of the differential equation $(\text{x+1})\frac{\text{dy}}{\text{dx}}=2\text{e}^{-\text{y}}-1,\ \text{given that y}=0\ \text{when x}=0.$
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Answer

Given: Differential equation $(\text{x+1})\frac{\text{dy}}{\text{dx}}=2\text{e}^{-\text{y}}-1$
$\Rightarrow\ \ (\text{x}+1)\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{e}^{\text{y}}}-1=\frac{2-\text{e}^{\text{y}}}{\text{e}^\text{y}}$ $\Rightarrow\ \ (\text{x}+1)\text{e}^\text{y}\ \text{dy}=(2-\text{e}^\text{y})\text{dx}$
$\Rightarrow\ \ \frac{\text{e}^\text{y}\text{dy}}{2-\text{e}^\text{y}}=\frac{\text{dx}}{\text{x}+1}$
$\text{Integrating both sides,}\ \ \int\frac{\text{e}^\text{y}}{2-\text{e}^\text{y}}\text{dy}=\int\frac{1}{\text{x}+1}\text{dx}$
$\text{Putting}\ \ \text{e}^\text{y}=\text{t}\ \ \Rightarrow\ \ \text{e}^\text{y}=\frac{\text{dt}}{\text{dy}}\ \ \Rightarrow\ \ \text{e}^\text{y}\ \text{dy}=\text{dt}$
$\therefore\ \ \int\frac{\text{dt}}{2-\text{t}}=\log|\text{x}+1|$ $\Rightarrow\ \ \frac{\log|2-\text{t}|}{-1}=\log|\text{x}+1|+\text{c}$
$\text{Putting e}^\text{y}=\text{t},\ \ -\log|2-\text{e}^\text{y}|=\log|\text{x}+1|+\text{c}$
$\Rightarrow\ \ \log|\text{x}+1|+\log|2-\text{e}^\text{y}|=-\text{c}$ $\Rightarrow\ \ \log|\text{x}+1||2-\text{e}^\text{y}|=-\text{c}$
$\Rightarrow\ \ |\text{x}+1||2-\text{e}^\text{y}|=-\text{c}$ $\Rightarrow\ \ (\text{x}+1)(2-\text{e}^\text{y})=\pm\text{e}^{-\text{c}}$
$\Rightarrow\ \ (\text{x}+1)(2-\text{e}^\text{y})=​​\text{C}\ \ \text{where C}=\pm\text{e}^{-\text{c}}\ \ ...\text{(i)}$
$\text{Putting x}=0,\ \text{y}=0\ \text{in eq. (i),} \ \ (1)(2-1)=\text{C}$ $\Rightarrow\ \ \text{C}=1$
$\text{Putting C}=1\ \text{in eq. (i),}\ \ (\text{x}+1)(2-\text{e}^\text{y})=1$
$\text{This solution may be written as}\ \ 2-\text{e}^\text{y}=\frac{1}{\text{x}+1}$ $\Rightarrow\ \ \text{e}^\text{y}=2-\frac{1}{\text{x}+1}=\frac{2\text{x}+1}{\text{x}+1}$
$\Rightarrow\ \ \log\text{e}^\text{y}=\log\Big(\frac{2\text{x}+1}{\text{x}+1}\Big)\ \ \Rightarrow\ \ \text{y}=\log\Big(\frac{2\text{x}+1}{\text{x}+1}\Big)$
where expresses y as an explicit function of x.

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