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Question: The solution of the D.E. \[x\cos ydy = (x.{e^x}\log x + {e^x})dx\] is A) \[\sin y = {e^x} + \log x...

The solution of the D.E. xcos⁡ydy=(x.exlog⁡x+ex)dxx\cos ydy = (x.{e^x}\log x + {e^x})dx is
A) sin⁡y=ex+log⁡x+c\sin y = {e^x} + \log x + c
B) sin⁡y−ex+log⁡x=c\sin y - {e^x} + \log x = c
C) sin⁡y=ex(log⁡x)+c\sin y = {e^x}(\log x) + c
D) None of these

Explanation

Solution

First shift from L.H.S. to R.H.S. because this will take all x terms to one side of the equation and then take ex{e^x}common. Here function and its derivative are on the same side.
Formula used:
∫cos⁡ydy=sin⁡y\int {\cos ydy = \sin y}
∫log⁡xdx=1x\int {\log xdx = \dfrac{1}{x}}

Complete step by step answer:

Given that,
xcos⁡ydy=(x.exlog⁡x+ex)dxx\cos ydy = (x.{e^x}\log x + {e^x})dx
Shift xxof L.H.S. to R.H.S

\Rightarrow \cos ydy = \left( {\dfrac{{x.{e^x}\log x + {e^x}}}{x}} \right)dx \\\ \Rightarrow \cos ydy = ({e^x}\log x + \dfrac{{{e^x}}}{x})dx \\\ \end{gathered} $$ Now integrating both sides, $$ \Rightarrow \int {\cos ydy = \int {({e^x}\log x + \dfrac{{{e^x}}}{x})dx} } $$ Taking $${e^x}$$ common $$ \Rightarrow \sin y = \int {{e^x}\left( {\log x + \dfrac{1}{x}} \right)dx} $$ Now ,here on R.H.S. we observe it is in the form of $$\int {{e^x}\left[ {f(x) + f'(x)} \right]} dx$$ We know that $$\int {{e^x}\left[ {f(x) + f'(x)} \right]} dx = {e^x}f(x) + c$$ $$ \Rightarrow \sin y = {e^x}\log x + c$$ Hence **option c is the correct answer.** **Note:** First carefully observe the sum because sometimes the answer is in the sum itself. Here logarithmic function and its derivative are on the same side of the sum. Don’t get confused about the derivative and integration of a function. **Additional information:** In mathematics, D.E. is an equation that relates one or more functions and their derivatives. Integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. If $$f(x)$$is a function then $$f'(x)$$is called its derivative. Derivative and integration are the two reverse processes of each other. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two.