Question

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

Hint: Use a variable separable method to solve the given differential equation. Integrate with respect to â€˜dxâ€™ and â€˜dyâ€™ to both sides and simplify it to get the solution of the given differential equation.

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 â€˜yâ€™ then we can separate variables â€˜xâ€™ and â€˜yâ€™ 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 â€˜xâ€™ to one side and functions of variable â€˜yâ€™ to another side to integrate the equation with respect to â€˜dxâ€™ and â€˜dyâ€™.

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

â€˜xâ€™ and â€˜yâ€™ 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 â€˜1â€™ 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 â€˜ln Câ€™, 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.