The solution of the differential equation
`(dy)/(dx)-(tany)/(x)=(tanysiny)/(x^(2))`, is

Options

A. `(x)/(siny)+logx-C`
B. `(y)/(sinx)+logx=C`
C. `logy+x=C`
D. `logx+y=C`

Answer

To solve the differential equation

\[ \frac{dy}{dx} - \frac{\tan y}{x} = \frac{\tan y \sin y}{x^2}, \]

we will follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the equation to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{\tan y}{x} + \frac{\tan y \sin y}{x^2}. \]

### Step 2: Dividing by \(\tan y \sin y\)
Next, we divide the entire equation by \(\tan y \sin y\) to simplify it:

\[ \frac{1}{\tan y \sin y} \frac{dy}{dx} - \frac{1}{x \sin y} = \frac{1}{x^2}. \]

### Step 3: Substituting \(t = \frac{1}{\sin y}\)
Now, we make the substitution \(t = \frac{1}{\sin y}\), which implies \(\sin y = \frac{1}{t}\). We differentiate this to find \(dy\):

\[ dy = -\frac{\cos y}{\sin^2 y} \, dy \Rightarrow dy = -\frac{\cos y}{\left(\frac{1}{t}\right)^2} \, dt = -t^2 \cos y \, dt. \]

### Step 4: Substitute into the Equation
Substituting \(dy\) into the equation gives us:

\[ -\frac{t^2 \cos y}{\tan y \sin^2 y} \frac{dt}{dx} - \frac{1}{x \sin y} = \frac{1}{x^2}. \]

### Step 5: Simplifying the Equation
Using the relationships \(\tan y = \frac{\sin y}{\cos y}\) and \(\sin y = \frac{1}{t}\), we can simplify the equation further.

### Step 6: Forming a Linear Differential Equation
We can rewrite the equation in the standard linear form:

\[ \frac{dt}{dx} + \frac{t}{x} = -\frac{1}{x^2}. \]

### Step 7: Finding the Integrating Factor
The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int \frac{1}{x} \, dx} = e^{\log x} = x. \]

### Step 8: Multiplying through by the Integrating Factor
Multiply the entire equation by the integrating factor:

\[ x \frac{dt}{dx} + t = -\frac{1}{x}. \]

### Step 9: Integrating Both Sides
Now we integrate both sides:

\[ \int \left( x \frac{dt}{dx} + t \right) dx = \int -\frac{1}{x} \, dx. \]

This gives us:

\[ xt = -\log x + C. \]

### Step 10: Substituting Back
Substituting back \(t = \frac{1}{\sin y}\):

\[ x \cdot \frac{1}{\sin y} = -\log x + C. \]

Rearranging gives us the final solution:

\[ \frac{x}{\sin y} + \log x = C. \]

### Final Answer
Thus, the solution of the differential equation is:

\[ \frac{x}{\sin y} + \log x = C. \]

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