Question

If $y = y\left( x \right)$ is the solution of the differential equation $\dfrac{{5 + {e^x}}}{{2 + y}}\dfrac{{dy}}{{dx}} + {e^x} = 0$ satisfying $y\left( 0 \right) = 1$,then find the value of $y\left( {{{\log }_e}13} \right)$.
A. 1
B. 0
C. 2
D. -1

Answer 1

Hint: First we will rewrite the differential equation in the form $f\left( y \right)dy = g\left( x \right)dx$. Then by using the substitution method, solve the equation. Then use the initial value to calculate the value of the integrating constant. Put $x = {\log _e}13$in the solution of the differential equation to calculate $y\left( {{{\log }_e}13} \right)$.

Formula Used:
$\int {\dfrac{1}{x}dx} = \log x + c$
$\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$

Complete step by step solution:
Given differential equation is $\dfrac{{5 + {e^x}}}{{2 + y}}\dfrac{{dy}}{{dx}} + {e^x} = 0$.
Subtract ${e^x}$ from both sides of the equation.
$ \Rightarrow \dfrac{{5 + {e^x}}}{{2 + y}}\dfrac{{dy}}{{dx}} + {e^x} - {e^x} = 0 - {e^x}$
$ \Rightarrow \dfrac{{5 + {e^x}}}{{2 + y}}\dfrac{{dy}}{{dx}} = - {e^x}$
Multiply $dx$ on both sides of the equation
$ \Rightarrow \dfrac{{\left( {5 + {e^x}} \right)}}{{\left( {2 + y} \right)}}dy = - {e^x}dx$
Divide both sides by $5 + {e^x}$.
$ \Rightarrow \dfrac{{dy}}{{\left( {2 + y} \right)}} = \dfrac{{ - {e^x}dx}}{{\left( {5 + {e^x}} \right)}}$
Take integration operations on both sides of the equation.
$ \Rightarrow \int {\dfrac{{dy}}{{\left( {2 + y} \right)}}} = \int {\dfrac{{ - {e^x}dx}}{{\left( {5 + {e^x}} \right)}}} $ …..(1)
Solve the integration $\int {\dfrac{{dy}}{{\left( {2 + y} \right)}}} $ by substitute method.
Let $2 + y = t$.
Differentiate both sides by $x$.
$ \Rightarrow dy = dt$
Substitute $2 + y = t$ and $dy = dt$ in $\int {\dfrac{{dy}}{{\left( {2 + y} \right)}}} $
$\int {\dfrac{{dy}}{{\left( {2 + y} \right)}}} $
$ = \int {\dfrac{{dt}}{t}} $
Apply the formula $\int {\dfrac{1}{x}dx} = {\log _e}x + c$
$ = {\log _e}t + {c_1}$
Substitute the value of $t$.
$ = {\log _e}\left| {2 + y} \right| + {c_1}$
Solve the integration $\int {\dfrac{{ - {e^x}dx}}{{\left( {5 + {e^x}} \right)}}} $ by substitute method.
Let $5 + {e^x} = z$.
Differentiate both sides by $x$.
$ \Rightarrow {e^x}dx = dz$
Substitute $5 + {e^x} = z$ and ${e^x}dx = dz$ in $\int {\dfrac{{ - {e^x}dx}}{{\left( {5 + {e^x}} \right)}}} $
$\int {\dfrac{{ - {e^x}dx}}{{\left( {5 + {e^x}} \right)}}} $
$ = - \int {\dfrac{{dz}}{z}} $
Apply the formula $\int {\dfrac{1}{x}dx} = {\log _e}x + c$
$ = - {\log _e}z + {c_2}$
Substitute the value of $z$.
$ = - {\log _e}\left| {5 + {e^x}} \right| + {c_2}$
Now putting $\int {\dfrac{{dy}}{{\left( {2 + y} \right)}}} = {\log _e}\left| {2 + y} \right| + {c_1}$ and $\int {\dfrac{{ - {e^x}dx}}{{\left( {5 + {e^x}} \right)}}} = - {\log _e}\left| {5 + {e^x}} \right| + {c_2}$in equation (1)
$\int {\dfrac{{dy}}{{\left( {2 + y} \right)}}} = \int {\dfrac{{ - {e^x}dx}}{{\left( {5 + {e^x}} \right)}}} $
$ \Rightarrow {\log _e}\left| {2 + y} \right| + {c_1} = - {\log _e}\left| {5 + {e^x}} \right| + {c_2}$
$ \Rightarrow {\log _e}\left| {2 + y} \right| + {\log _e}\left| {5 + {e^x}} \right| = {c_2} - {c_1}$
Apply the formula of sum logarithm
$ \Rightarrow {\log _e}\left| {\left( {2 + y} \right)\left( {5 + {e^x}} \right)} \right| = c$
Apply the inverse of the logarithm:
$ \Rightarrow \left( {2 + y} \right)\left( {5 + {e^x}} \right) = {e^c}$ …..(2)
Now put $x = 0$ and $y = 1$ in the above equation.
$ \Rightarrow \left( {2 + 1} \right)\left( {5 + {e^0}} \right) = {e^c}$
Substitute ${e^0} = 1$.
$ \Rightarrow \left( {2 + 1} \right)\left( {5 + 1} \right) = {e^c}$
$ \Rightarrow 18 = {e^c}$
Substitute the value of ${e^c}$ in the equation (2)
$\therefore \left( {2 + y} \right)\left( {5 + {e^x}} \right) = 18$
To calculate $y\left( {{{\log }_e}13} \right)$, substitute $x = {\log _e}13$ in the above equation.
$\therefore \left( {2 + y} \right)\left( {5 + {e^{{{\log }_e}13}}} \right) = 18$
$ \Rightarrow \left( {2 + y} \right)\left( {5 + 13} \right) = 18$
$ \Rightarrow \left( {2 + y} \right) \cdot 18 = 18$
Divide both sides by 18
$ \Rightarrow \left( {2 + y} \right) = 1$
Subtract 2 from both sides
$ \Rightarrow y = 1 - 2$
$ \Rightarrow y = - 1$

Option ‘D’ is correct

Note: If a differential equation is a mixture of two variables, then we have to rewrite the equation such that each side of the equation must contain only variables. Then we apply the substitution method.
Substitution method is a method to convert the given integration to the simplest form by substituting the independent variable with others.
Remember to find the integration constant, we need to put the initial value in the solution of the differential equation.