Question

What is the solution to the differential equation\[\dfrac{{dy}}{{dx}} = y\left( {{e^x} + 1} \right)\]?
A. \[y + {e^{\left( {{e^x} + x + c} \right)}} = 0\]
B. \[\log y = {e^x} + x + c\]
C. \[\log y + {e^x} = x + c\]
D. None of these

Answer 1

Hint: We will use the separation of variable method to separate the variable of the given differential equation. Then we take integration on both sides of the equation and integrate it.

Formula used:
Integration exponential function:
\[\int {{e^x}dx} = {e^x} + c\]
Integrating formula of logarithm
\[\int {\dfrac{1}{x}} dx = \log x + c\]
Integration formula
\[\int {dx} = x + c\]

Complete step by step solution:
The given differential equation is
\[\dfrac{{dy}}{{dx}} = y\left( {{e^x} + 1} \right)\]
Separate the variable of the above equation
\[\dfrac{{dy}}{y} = \left( {{e^x} + 1} \right)dx\]
Taking integration on both sides
\[\int {\dfrac{{dy}}{y}} = \int {\left( {{e^x} + 1} \right)dx} \]
Applying the integration formula
\[\log y = {e^x} + x + c\]
Hence option B is the correct option.

Additional information:
Particular solution: when we have an initial condition to calculate the value of c, then the solution of the differential equation is known as a particular solution.
General solution: When we don’t have any initial condition to calculate the value of c, then the solution of the differential equation is known as a general solution.
The general solution represents a family of solutions of the differential equation.

Note: Some students make a mistake to integrate \[\int {\dfrac{1}{y}} dy\]. They apply exponential formula that is \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\]. They calculate \[\int {\dfrac{1}{y}} dy = \dfrac{{{y^{ - 1 + 1}}}}{{ - 1 + 1}} + c\] which is an incorrect way. The correct formula is \[\int {\dfrac{1}{y}} dy = \log y + c\].