Question

Choose the solution of \[x\cos ydy = \left( {x{e^x}\log x + {e^x}} \right)dx\] from the following options.
A \[\sin y = \dfrac{1}{x}{e^x} + c\]
B \[\sin y + {e^x}\log x + c = 0\]
C \[\sin y = {e^x}\log x + c\]
D None of these.

Answer 1

Hint: Variable separation method is one of the methods of finding a general solution of the differential equation. There are x and y variables in the given differential equation. Separate the terms containing x and y. Then integrate it and simplify to find a general solution.

Formula used:
\[\int {\cos ydy} = \sin y + c\]
Integration by parts,
\[\int {f(x)g'(x)dx} = f(x)g(x) - \int {f'(x)g(x)dx} \]
Where, c is an arbitrary constant.

Complete step by step solution:
The given differential equation is \[x\cos ydy = \left( {x{e^x}\log x + {e^x}} \right)dx\].
Divide the equation by \[x\] to simplify the equation.
\[\begin{array}{l}\dfrac{{x\cos y}}{x}dy = \left( {\dfrac{{x{e^x}\log x + {e^x}}}{x}} \right)dx\\\cos ydy = \left( {{e^x}\log x + \dfrac{{{e^x}}}{x}} \right)dx\end{array}\]
Now integrate the equation.
\[\begin{array}{l}\int {\cos ydy} = \int {\left( {{e^x}\log x + \dfrac{{{e^x}}}{x}} \right)dx} \\\sin y = \int {{e^x}\log xdx} + \int {\dfrac{{{e^x}}}{x}dx} \end{array}\]
Use integration by parts to simplify further.
\[\begin{array}{l}\sin y = {e^x}\log x - \int {\dfrac{{{e^x}}}{x}dx + \int {\dfrac{{{e^x}}}{x}dx} } \\\sin y = {e^x}\log x + c\end{array}\]
So, the general solution of the differential equation is \[\sin y = {e^x}\log x + c\].
Hence option C is the correct option.

Note:The common mistake done here is solving the term \[ - \int {\dfrac{{{e^x}}}{x}} dx\]. In this it is not required to integrate it as we have \[\int {\dfrac{{{e^x}}}{x}} dx\]. It will get canceled. Sometimes students make mistakes while choosing f(x) and g(x) in integration by parts method. To choose f(x) we follow ILATE rule.