Question

The solution of the differential equation \( xdy - ydx = (x^2+y^2)dx \), is

• A. \( y = \tan(x+c) \); where c is an arbitrary constant
• B. \( x = y \tan(x+c) \); where c is an arbitrary constant
• C. \( y = x \tan^{-1}(y+c) \); where c is an arbitrary constant
• D. \( y = x \tan(x+c) \); where c is an arbitrary constant

Hint:

When you see the combination \(xdy - ydx\), immediately think about dividing by \(x^2\), \(y^2\), or \(x^2+y^2\). These lead to the differentials of \(y/x\), \(-x/y\), and \(\tan^{-1}(y/x)\) respectively, which are key to solving many first-order ODEs.

Answer 1

The Correct Option is D

Step 1: Problem Definition:
This is a first-order differential equation. Its structure hints at simplification through rearrangement and recognition of a standard differential form.

Step 2: Strategy:
The expression \( \frac{xdy - ydx}{x^2} \) is the differential of \( \frac{y}{x} \), and \( \frac{xdy - ydx}{x^2+y^2} \) is the differential of \( \tan^{-1}(\frac{y}{x}) \). The approach is to rearrange the equation to isolate one of these forms. Rearrange the equation: \[ xdy - ydx = (x^2+y^2)dx \]Divide by \( x^2 \):\[ \frac{xdy - ydx}{x^2} = \left(1 + \frac{y^2}{x^2}\right)dx \]The left side is \( d\left(\frac{y}{x}\right) \).\[ d\left(\frac{y}{x}\right) = \left(1 + \left(\frac{y}{x}\right)^2\right)dx \]
Step 3: Solution Steps:
Let \( v = \frac{y}{x} \). This transforms the equation into a separable differential equation in terms of \(v\) and \(x\):\[ dv = (1+v^2)dx \]Separate the variables:\[ \frac{dv}{1+v^2} = dx \]Integrate both sides:\[ \int \frac{1}{1+v^2} dv = \int dx \]\[ \tan^{-1}(v) = x + c \]where c is the constant of integration.Substitute back \( v = \frac{y}{x} \):\[ \tan^{-1}\left(\frac{y}{x}\right) = x + c \]To match solution formats, take the tangent of both sides:\[ \frac{y}{x} = \tan(x+c) \]\[ y = x \tan(x+c) \]
Step 4: Answer:
The solution is \( y = x \tan(x+c) \).