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.