# Solve the differential equation $\left( {{e}^{x}}+1 \right)ydy+\left( y+1 \right)dx=0$.

Last updated date: 25th Mar 2023

•

Total views: 306.3k

•

Views today: 2.83k

Answer

Verified

306.3k+ views

Hint: Use a variable separable method to solve the given differential equation. Integrate with respect to ‘dx’ and ‘dy’ to both sides and simplify it to get the solution of the given differential equation.

Complete step-by-step answer:

We have

$\left( {{e}^{x}}+1 \right)ydy+\left( y+1 \right)dx=0.............\left( i \right)$

Dividing the whole equation by dx, we get,

$\left( {{e}^{x}}+1 \right)y\dfrac{dy}{dx}+\left( y+1 \right)\dfrac{dx}{dx}=0$

Or

$\left( {{e}^{x}}+1 \right)y\dfrac{dy}{dx}+\left( y+1 \right)=0..........(ii)$

Now, as we know, we have three types of differential equations i.e., separable, linear and homogeneous.

Now, by observation, we get that if we divide equation (ii), by ‘y’ then we can separate variables ‘x’ and ‘y’ easily. Hence, the given differential equation belongs to a separable type.

So, on dividing equation (ii), we get

$\left( {{e}^{x}}+1 \right)\dfrac{y}{y}\dfrac{dy}{dx}+\left( \dfrac{y+1}{y} \right)=0$

Or

$\left( {{e}^{x}}+1 \right)\dfrac{dy}{dx}+\dfrac{y+1}{y}=0$

Now transferring $\dfrac{y+1}{y}$ to other side, we get

$\left( {{e}^{x}}+1 \right)\dfrac{dy}{dx}=-\dfrac{\left( y+1 \right)}{y}..........(iii)$

Now, we can transfer functions of variable ‘x’ to one side and functions of variable ‘y’ to another side to integrate the equation with respect to ‘dx’ and ‘dy’.

So, equation (iii) can be written as

$\left( \dfrac{y}{y+1} \right)dy=\dfrac{-1}{{{e}^{x}}+1}dx$

Now, we observe that variable are easily separated, so we can integrate them with respect to

‘x’ and ‘y’ hence, we get

$\int{\dfrac{y}{y+1}dy=-\int{\dfrac{1}{{{e}^{x}}+1}dx............(iv)}}$

Let ${{I}_{1}}=\int{\dfrac{y}{y+1}dy}$ and ${{I}_{2}}=\int{\dfrac{1}{{{e}^{x}}+1}dx}$

Let us solve both the integration individually.

So, we have ${{I}_{1}}$ as,

${{I}_{1}}=\int{\dfrac{y}{y+1}dy}$

Adding and subtracting ‘1’ in numerator, we get,

${{I}_{1}}=\int{\dfrac{\left( y+1 \right)-1}{\left( y+1 \right)}dy}$

Now, we can separate (y+1) as

\[{{I}_{1}}=\int{\dfrac{y+1}{y+1}dy}-\int{\dfrac{1}{y+1}dy}\]

Or

\[{{I}_{1}}=\int{1dy}-\int{\dfrac{1}{y+1}dy}\]

As we know,

$\begin{align}

& \int{\dfrac{1}{x}dx}=\ln x \\

& \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}} \\

\end{align}$

So, ${{I}_{1}}$ can be simplified as,

${{I}_{1}}=\int{{{y}^{0}}dy-\ln \left( y+1 \right)+{{C}_{1}}}$

\[{{I}_{1}}=y-\ln \left( y+1 \right)+{{C}_{1}}............\left( v \right)\]

Now, we have ${{I}_{2}}$ as

${{I}_{2}}=\int{\dfrac{1}{{{e}^{x}}+1}dx}$

Multiplying by ${{e}^{-x}}$ in numerator and denominator we get,

${{I}_{2}}=\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}.{{e}^{-x}}+{{e}^{-x}}}dx}$

As we have property of surds as, ${{m}^{a}}.{{m}^{b}}={{m}^{a+b}}$

So, we can write ${{I}_{2}}$ as

${{I}_{2}}=\int{\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}dx}..............\left( vi \right)$

Let us suppose $1+{{e}^{-x}}=t$.

Differentiating both sides w.r.t. x, we get

$-{{e}^{-x}}=\dfrac{dt}{dx}$

Where $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$, so, we have

${{e}^{-x}}dx=-dt$

Substituting these values in equation (vi), we get

${{I}_{2}}=\int{\dfrac{-dt}{t}=-\int{\dfrac{dt}{t}}}$

Now, we know that $\int{\dfrac{1}{t}dt=\ln t}$ hence,

${{I}_{2}}=-\ln t+{{C}_{2}}$

Since, we have value of t as \[1+{{e}^{-x}}\], hence ${{I}_{2}}$ in terms of x can be given as

${{I}_{2}}=-\ln \left( 1+{{e}^{-x}} \right)+{{C}_{2}}.............\left( vii \right)$

Hence, from equation (iv), (v) and (vii) we get,

$\begin{align}

& y-\ln \left( y+1 \right)+{{C}_{1}}=-\left( -\ln \left( 1+{{e}^{-x}} \right)+{{C}_{2}} \right) \\

& y-\ln \left( y+1 \right)+{{C}_{1}}=\ln \left( 1+{{e}^{-x}} \right)-{{C}_{2}} \\

& y-\ln \left( y+1 \right)=\ln \left( 1+{{e}^{-x}} \right)+{{C}_{3}} \\

\end{align}$

Where $-{{C}_{2}}-{{C}_{1}}={{C}_{3}}$

Let us replace ${{C}_{3}}$ by ‘ln C’, so we get above equation as

$y-\ln \left( y+1 \right)=\ln \left( 1+{{e}^{-x}} \right)+\ln C........\left( viii \right)$

Now, we know that \[\text{ln }a+\ln b=\ln ab\], so, equation (viii) can be given as

$y=\ln \left( y+1 \right)+\ln \left( C\left( 1+{{e}^{-x}} \right) \right)$

$y=\ln \left( C\left( y+1 \right)\left( 1+{{e}^{-x}} \right) \right)$

As we know that if ${{a}^{x}}=N$ then $x={{\log }_{a}}N$ or vice versa.

Hence, above equation can be written as

$C\left( y+1 \right)\left( 1+{{e}^{-x}} \right)={{e}^{y}}$

This is the required solution.

Note: One can go wrong if trying to solve the given differential equation by a homogenous method, as the given equation is not a homogeneous differential equation so we cannot apply this method to a given variable separable differential equation.

Observing the given differential equation as a variable separable equation is the key point of the question.

Complete step-by-step answer:

We have

$\left( {{e}^{x}}+1 \right)ydy+\left( y+1 \right)dx=0.............\left( i \right)$

Dividing the whole equation by dx, we get,

$\left( {{e}^{x}}+1 \right)y\dfrac{dy}{dx}+\left( y+1 \right)\dfrac{dx}{dx}=0$

Or

$\left( {{e}^{x}}+1 \right)y\dfrac{dy}{dx}+\left( y+1 \right)=0..........(ii)$

Now, as we know, we have three types of differential equations i.e., separable, linear and homogeneous.

Now, by observation, we get that if we divide equation (ii), by ‘y’ then we can separate variables ‘x’ and ‘y’ easily. Hence, the given differential equation belongs to a separable type.

So, on dividing equation (ii), we get

$\left( {{e}^{x}}+1 \right)\dfrac{y}{y}\dfrac{dy}{dx}+\left( \dfrac{y+1}{y} \right)=0$

Or

$\left( {{e}^{x}}+1 \right)\dfrac{dy}{dx}+\dfrac{y+1}{y}=0$

Now transferring $\dfrac{y+1}{y}$ to other side, we get

$\left( {{e}^{x}}+1 \right)\dfrac{dy}{dx}=-\dfrac{\left( y+1 \right)}{y}..........(iii)$

Now, we can transfer functions of variable ‘x’ to one side and functions of variable ‘y’ to another side to integrate the equation with respect to ‘dx’ and ‘dy’.

So, equation (iii) can be written as

$\left( \dfrac{y}{y+1} \right)dy=\dfrac{-1}{{{e}^{x}}+1}dx$

Now, we observe that variable are easily separated, so we can integrate them with respect to

‘x’ and ‘y’ hence, we get

$\int{\dfrac{y}{y+1}dy=-\int{\dfrac{1}{{{e}^{x}}+1}dx............(iv)}}$

Let ${{I}_{1}}=\int{\dfrac{y}{y+1}dy}$ and ${{I}_{2}}=\int{\dfrac{1}{{{e}^{x}}+1}dx}$

Let us solve both the integration individually.

So, we have ${{I}_{1}}$ as,

${{I}_{1}}=\int{\dfrac{y}{y+1}dy}$

Adding and subtracting ‘1’ in numerator, we get,

${{I}_{1}}=\int{\dfrac{\left( y+1 \right)-1}{\left( y+1 \right)}dy}$

Now, we can separate (y+1) as

\[{{I}_{1}}=\int{\dfrac{y+1}{y+1}dy}-\int{\dfrac{1}{y+1}dy}\]

Or

\[{{I}_{1}}=\int{1dy}-\int{\dfrac{1}{y+1}dy}\]

As we know,

$\begin{align}

& \int{\dfrac{1}{x}dx}=\ln x \\

& \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}} \\

\end{align}$

So, ${{I}_{1}}$ can be simplified as,

${{I}_{1}}=\int{{{y}^{0}}dy-\ln \left( y+1 \right)+{{C}_{1}}}$

\[{{I}_{1}}=y-\ln \left( y+1 \right)+{{C}_{1}}............\left( v \right)\]

Now, we have ${{I}_{2}}$ as

${{I}_{2}}=\int{\dfrac{1}{{{e}^{x}}+1}dx}$

Multiplying by ${{e}^{-x}}$ in numerator and denominator we get,

${{I}_{2}}=\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}.{{e}^{-x}}+{{e}^{-x}}}dx}$

As we have property of surds as, ${{m}^{a}}.{{m}^{b}}={{m}^{a+b}}$

So, we can write ${{I}_{2}}$ as

${{I}_{2}}=\int{\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}dx}..............\left( vi \right)$

Let us suppose $1+{{e}^{-x}}=t$.

Differentiating both sides w.r.t. x, we get

$-{{e}^{-x}}=\dfrac{dt}{dx}$

Where $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$, so, we have

${{e}^{-x}}dx=-dt$

Substituting these values in equation (vi), we get

${{I}_{2}}=\int{\dfrac{-dt}{t}=-\int{\dfrac{dt}{t}}}$

Now, we know that $\int{\dfrac{1}{t}dt=\ln t}$ hence,

${{I}_{2}}=-\ln t+{{C}_{2}}$

Since, we have value of t as \[1+{{e}^{-x}}\], hence ${{I}_{2}}$ in terms of x can be given as

${{I}_{2}}=-\ln \left( 1+{{e}^{-x}} \right)+{{C}_{2}}.............\left( vii \right)$

Hence, from equation (iv), (v) and (vii) we get,

$\begin{align}

& y-\ln \left( y+1 \right)+{{C}_{1}}=-\left( -\ln \left( 1+{{e}^{-x}} \right)+{{C}_{2}} \right) \\

& y-\ln \left( y+1 \right)+{{C}_{1}}=\ln \left( 1+{{e}^{-x}} \right)-{{C}_{2}} \\

& y-\ln \left( y+1 \right)=\ln \left( 1+{{e}^{-x}} \right)+{{C}_{3}} \\

\end{align}$

Where $-{{C}_{2}}-{{C}_{1}}={{C}_{3}}$

Let us replace ${{C}_{3}}$ by ‘ln C’, so we get above equation as

$y-\ln \left( y+1 \right)=\ln \left( 1+{{e}^{-x}} \right)+\ln C........\left( viii \right)$

Now, we know that \[\text{ln }a+\ln b=\ln ab\], so, equation (viii) can be given as

$y=\ln \left( y+1 \right)+\ln \left( C\left( 1+{{e}^{-x}} \right) \right)$

$y=\ln \left( C\left( y+1 \right)\left( 1+{{e}^{-x}} \right) \right)$

As we know that if ${{a}^{x}}=N$ then $x={{\log }_{a}}N$ or vice versa.

Hence, above equation can be written as

$C\left( y+1 \right)\left( 1+{{e}^{-x}} \right)={{e}^{y}}$

This is the required solution.

Note: One can go wrong if trying to solve the given differential equation by a homogenous method, as the given equation is not a homogeneous differential equation so we cannot apply this method to a given variable separable differential equation.

Observing the given differential equation as a variable separable equation is the key point of the question.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main