Question

The slope of a curve at any point \[\left( {x,y} \right)\] other than the origin, is \[y + \dfrac{y}{x}\]. Then, the equation of the curve is:
A. \[y = Cx{e^x}\]
B. \[y = x\left( {{e^x} + C} \right)\]
C. \[xy = C{e^x}\]
D. \[y + x{e^x} = C\]
E. \[\left( {y - x} \right){e^x} = C\]

Answer 1

Hint: First of all, equate the given slope with \[\dfrac{{dy}}{{dx}}\]. Now, write the obtained equation in the form of \[g\left( y \right)dy = f\left( x \right)dx\] and then integrate left side of the equation with respect to \[y\] and right side of the equation with respect to \[x\]. As we know that, \[\ln {e^x} = x = {e^{\ln x}}\] , so we can use this to convert the expression in simplified form.

Complete step by step solution:
Consider the given expression, \[y + \dfrac{y}{x}\].
We have to find the equation of a curve which has slope equals to \[y + \dfrac{y}{x}\].
We know that the slope of any curve \[ = \dfrac{{dy}}{{dx}}\]. In this question slope is given as \[y + \dfrac{y}{x}\].
Therefore, we can say that \[\dfrac{{dy}}{{dx}} = y + \dfrac{y}{x}\].
Also, this slope is defined for all \[\left( {x,y} \right)\] other than \[\left( {0,0} \right)\].
Therefore, we have \[\dfrac{{dy}}{{dx}} = y + \dfrac{y}{x}\] where \[\left( {x,y} \right) \ne \left( {0,0} \right)\].

Now, we need to solve the equation \[\dfrac{{dy}}{{dx}} = y + \dfrac{y}{x}\],
In order to solve this differential equation, rewrite this equation in the form of \[g\left( y \right)dy = f\left( x \right)dx\]
Thus, we get,
\[
   \Rightarrow \dfrac{{dy}}{{dx}} = y + \dfrac{y}{x} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = y\left( {1 + \dfrac{1}{x}} \right) \\
   \Rightarrow \left( {\dfrac{1}{y}} \right)dy = \left( {1 + \dfrac{1}{x}} \right)dx \\
 \]

We know that \[\left( 1 \right) \cdot dx = x + C\] and \[\dfrac{1}{x}dx = \ln \left| x \right| + \ln C\] where \[C\] is constant of integration.
So, integrate left side of the equation \[\left( {\dfrac{1}{y}} \right)dy = \left( {1 + \dfrac{1}{x}} \right)dx\] with respect to \[y\] and right side of the equation with respect to \[x\].

\[
   \Rightarrow \int {\left( {\dfrac{1}{y}} \right)dy} = \int {\left( {1 + \dfrac{1}{x}} \right)dx} \\
   \Rightarrow \ln \left| y \right| = \int {\left( 1 \right)dx} + \int {\left( {\dfrac{1}{x}} \right)dx} \\
   \Rightarrow \ln \left| y \right| = x + \ln \left| x \right| + \ln C \\
 \]

We know that natural logarithm function is the inverse function of exponential function and vice versa. That is, for any \[x \in \mathbb{R}\], \[\ln {e^x} = x = {e^{\ln x}}\] equation(i)
Rewrite \[x\] as \[\ln {e^x}\] using equation (i) in the equation obtained above that is \[\ln \left| y \right| = x + \ln \left| x \right| + \ln C\] and simply the equation using the multiplication property of logarithmic function.
Thus, we have,
\[
   \Rightarrow \ln \left| y \right| = \ln {e^x} + \ln \left| x \right| + \ln C \\
   \Rightarrow \ln \left| y \right| = \ln \left| {Cx{e^x}} \right| \\
   \Rightarrow y = {e^{\ln \left| {Cx{e^x}} \right|}} \\
   \Rightarrow y = Cx{e^x} \\
 \]
Therefore, the correct option is A.

Note: In this question, first of all, note that the slope is given for all points \[\left( {x,y} \right)\] other than the origin in the form of an expression. Slope is not a constant value. So, we have to solve this question using differentiation and integration concepts.