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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS1 Mark
Question
Choose the correct answer from the given four options.The general solution of the differential equation $(\text{e}^{\text{x}}+1)\text{ydy}=(\text{y}+1)\text{e}^{\text{x}}$ is:
  1. $(\text{y}+1)=\text{k}(\text{e}^{\text{x}}+1)$
  2. $\text{y}+1=\text{e}^{\text{x}}+1+\text{k}$
  3. $\text{y}=\log\left\{\text{k}(\text{y}+1)(\text{e}^{\text{x}}+1)\right\}$
  4. $\text{y}=\log\left\{\frac{\text{e}^{\text{x}}+1}{\text{y}+1}\right\}+\text{k}$
✓

Answer

  1. $\text{y}=\log\left\{\text{k}(\text{y}+1)(\text{e}^{\text{x}}+1)\right\}$
Solution:
Given differential equation
$(\text{e}^{\text{x}}+1)\text{ydy}=(\text{y}+1)\text{e}^{\text{x}}\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}(1+\text{y})}{(\text{e}^{\text{x}}+1)\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{e}^{\text{x}}+1)\text{y}}{\text{e}^{\text{x}}(1+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}\text{y}}{\text{e}^{\text{x}}(1+\text{y})}\frac{\text{y}}{\text{e}^{\text{x}}(1+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}+\frac{\text{y}}{(1+\text{y})\text{e}^{\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(1+\frac{1}{\text{e}^{\text{x}}}\Big)\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(1+\frac{1}{\text{e}^{\text{x}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(\frac{\text{e}^{\text{x}}+1}{\text{e}^{\text{x}}}\Big)$
$\Rightarrow\Big(\frac{\text{y}}{1+\text{y}}\Big)\text{dy}=\Big(\frac{\text{e}^{\text{x}}}{\text{e}^{\text{x}}+1}\Big)\text{dx}$
On integrating both sides, we get
$\int\frac{\text{y}}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\int\frac{1+\text{y}-1}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\int1\text{dy}-\int\frac{\text{y}}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\text{y}-\log|(1+\text{y})+\log(1+\text{e}^{\text{x}})|+\log(\text{k})$
$\Rightarrow\text{y}=\log\left\{\text{k}(1+\text{y})(1+\text{e}^{\text{x}})\right\}$

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