# The solution of differential equation $({{e}^{x}}+1)ydy=(y+1){{e}^{x}}dx$ is,

A. $({{e}^{x}}+1)(y+1)=c{{e}^{y}}$

B. $({{e}^{x}}+1)\left| y+1 \right|=c{{e}^{-y}}$

C. $({{e}^{x}}+1)(y+1)=\pm c{{e}^{y}}$

D. None of these

Answer

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Hint: First of all in any differential equation try to separate the variable with their respective differential and then start further solving by method of variable separable form of differential equation, otherwise think of another method that you already know to solve it.

Complete step-by-step answer:

We will separate the variable and try to solve it so, let’s begin with that and we will get;

\[\Rightarrow \dfrac{y}{y+1}dy=\dfrac{{{e}^{x}}}{{{e}^{x}}+1}dx\]

Now, further we will integrate both sides and we get;

\[\Rightarrow \int{\dfrac{y}{y+1}}dy=\int{\dfrac{{{e}^{x}}}{{{e}^{x}}+1}dx}\]

\[\Rightarrow \int{1-\dfrac{1}{1+y}}dy=\int{\dfrac{{{e}^{x}}}{{{e}^{x}}+1}}dx\].................................(a)

Now, we will integrate the both sides of the equality separately as shown below;

\[\Rightarrow \int{1-\dfrac{1}{y+1}}dy=y-\ln \left| y+1 \right|\text{ + }\ln c\text{ }......................\text{(1)}\]

Also, we have $\int{\dfrac{{{e}^{x}}}{{{e}^{x}}+1}dx}$.

Here, let’s suppose that $1+{{e}^{x}}=u$.

$\Rightarrow {{e}^{x}}dx=du$.

Now, we will substitute these values in above integral and we get :

$\Rightarrow \int{\dfrac{1}{u}du=\ln u=\ln (1+{{e}^{x}})...................(2)}$

Now, we will substitute the above value of integrals (1) and (2) in the equation (a) and we get;

\[\begin{align}

& y-\ln \left| y+1 \right|+\ln c=\ln (1+{{e}^{x}}) \\

& \Rightarrow y=\ln (1+{{e}^{x}})+\ln \left| y+1 \right|-\ln c \\

\end{align}\]

Here, we will use the logarithmic formulae which are given below.

These formulae are \[\ln a+\ln b=\ln (ab)\text{ }\]and $\ln a-\ln b=\ln (\dfrac{a}{b})$.

So, on further simplification we will get,

\[\begin{align}

& \Rightarrow y=\ln \dfrac{(1+{{e}^{x}})\left| 1+y \right|}{c} \\

& \Rightarrow {{e}^{y}}=\dfrac{(1+{{e}^{x}})\left| 1+y \right|}{c} \\

& \Rightarrow c{{e}^{y}}=(1+{{e}^{x}})\left| 1+y \right| \\

& \Rightarrow \pm c{{e}^{y}}=(1+{{e}^{x}})(1+y) \\

\end{align}\]

Hence, the solution of the given differential equation is \[({{e}^{x}}+1)(1+y)=\pm c{{e}^{y}}\].

Therefore, the correct option of the above question will be C.

NOTE:

Be careful while doing calculations because there are many places where you can make a mistake. Take care of signs during calculation since it will also change your final answer.

Also take care of the modulus sign which is very important and if you will miss it then the whole solution will become wrong and you will get the incorrect option.

Complete step-by-step answer:

We will separate the variable and try to solve it so, let’s begin with that and we will get;

\[\Rightarrow \dfrac{y}{y+1}dy=\dfrac{{{e}^{x}}}{{{e}^{x}}+1}dx\]

Now, further we will integrate both sides and we get;

\[\Rightarrow \int{\dfrac{y}{y+1}}dy=\int{\dfrac{{{e}^{x}}}{{{e}^{x}}+1}dx}\]

\[\Rightarrow \int{1-\dfrac{1}{1+y}}dy=\int{\dfrac{{{e}^{x}}}{{{e}^{x}}+1}}dx\].................................(a)

Now, we will integrate the both sides of the equality separately as shown below;

\[\Rightarrow \int{1-\dfrac{1}{y+1}}dy=y-\ln \left| y+1 \right|\text{ + }\ln c\text{ }......................\text{(1)}\]

Also, we have $\int{\dfrac{{{e}^{x}}}{{{e}^{x}}+1}dx}$.

Here, let’s suppose that $1+{{e}^{x}}=u$.

$\Rightarrow {{e}^{x}}dx=du$.

Now, we will substitute these values in above integral and we get :

$\Rightarrow \int{\dfrac{1}{u}du=\ln u=\ln (1+{{e}^{x}})...................(2)}$

Now, we will substitute the above value of integrals (1) and (2) in the equation (a) and we get;

\[\begin{align}

& y-\ln \left| y+1 \right|+\ln c=\ln (1+{{e}^{x}}) \\

& \Rightarrow y=\ln (1+{{e}^{x}})+\ln \left| y+1 \right|-\ln c \\

\end{align}\]

Here, we will use the logarithmic formulae which are given below.

These formulae are \[\ln a+\ln b=\ln (ab)\text{ }\]and $\ln a-\ln b=\ln (\dfrac{a}{b})$.

So, on further simplification we will get,

\[\begin{align}

& \Rightarrow y=\ln \dfrac{(1+{{e}^{x}})\left| 1+y \right|}{c} \\

& \Rightarrow {{e}^{y}}=\dfrac{(1+{{e}^{x}})\left| 1+y \right|}{c} \\

& \Rightarrow c{{e}^{y}}=(1+{{e}^{x}})\left| 1+y \right| \\

& \Rightarrow \pm c{{e}^{y}}=(1+{{e}^{x}})(1+y) \\

\end{align}\]

Hence, the solution of the given differential equation is \[({{e}^{x}}+1)(1+y)=\pm c{{e}^{y}}\].

Therefore, the correct option of the above question will be C.

NOTE:

Be careful while doing calculations because there are many places where you can make a mistake. Take care of signs during calculation since it will also change your final answer.

Also take care of the modulus sign which is very important and if you will miss it then the whole solution will become wrong and you will get the incorrect option.

Last updated date: 21st Sep 2023

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