Question

If xdy - ydx = \(\sqrt{x^2 + y^2}\) dx. If y = y(x) \(\&\) y(1) = 0 then y(3) is :

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Recognize homogeneous differential equations of the form \(\frac{dy}{dx} = f\left(\frac{y}{x}\right)\). The standard substitution \(y = vx\) simplifies them to a separable form, making them easier to solve.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a first-order differential equation and an initial condition. The task is to find the value of the function \(y\) at \(x = 3\). The equation is a homogeneous differential equation.

Step 2: Key Formula or Approach:
First, we rearrange the given equation into the standard form: \[ x \frac{dy}{dx} = y + \sqrt{x^2 + y^2} \] \[ \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x} = \frac{y}{x} + \sqrt{1 + \left( \frac{y}{x} \right)^2} \] This is a homogeneous differential equation, and we use the substitution \( y = vx \), which implies that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). 

Step 3: Detailed Explanation:
By substituting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) into the equation, we get: \[ v + x \frac{dv}{dx} = v + \sqrt{1 + v^2} \] \[ x \frac{dv}{dx} = \sqrt{1 + v^2} \] Now we separate the variables: \[ \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} \] Integrating both sides: \[ \int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x} \] Using the standard integral \( \int \frac{1}{\sqrt{a^2 + x^2}} dx = \ln|x + \sqrt{a^2 + x^2}| \), where \(a = 1\), we have: \[ \ln|v + \sqrt{1 + v^2}| = \ln|x| + C \] Rewriting the constant \(C\) as \( \ln|A| \), we get: \[ \ln|v + \sqrt{1 + v^2}| = \ln|Ax| \] \[ v + \sqrt{1 + v^2} = Ax \] Substituting back \( v = \frac{y}{x} \): \[ \frac{y}{x} + \sqrt{1 + \left( \frac{y}{x} \right)^2} = Ax \] Assuming \( x>0 \) (since the initial condition is at \( x = 1 \)), we get: \[ \frac{y + \sqrt{x^2 + y^2}}{x} = Ax \] \[ y + \sqrt{x^2 + y^2} = Ax^2 \] Applying the initial condition \( y(1) = 0 \): \[ 0 + \sqrt{1^2 + 0^2} = A(1)^2 \] \[ 1 = A \] The particular solution is: \[ y + \sqrt{x^2 + y^2} = x^2 \] To find \( y(3) \), substitute \( x = 3 \): \[ y(3) + \sqrt{3^2 + (y(3))^2} = 9 \] Let \( y(3) = y \): \[ y + \sqrt{9 + y^2} = 9 \] \[ \sqrt{9 + y^2} = 9 - y \] Squaring both sides: \[ 9 + y^2 = (9 - y)^2 = 81 - 18y + y^2 \] \[ 9 = 81 - 18y \] \[ 18y = 72 \] \[ y = 4 \] 
Step 4: Final Answer:
The value of \( y(3) \) is 4.