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

Answer

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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.

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