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. We need to find the value of the function y at x=3. The given equation is a homogeneous differential equation. 

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

Step 3: Detailed Explanation: 
Substitute \(y = vx\) and \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) into the equation: \[ v + x\frac{dv}{dx} = v + \sqrt{1 + v^2} \] \[ x\frac{dv}{dx} = \sqrt{1 + v^2} \] Now, we separate the variables: \[ \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} \] Integrate both sides: \[ \int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x} \] The standard integral \(\int \frac{1}{\sqrt{a^2+x^2}}dx = \ln|x + \sqrt{a^2+x^2}|\). Here \(a=1\). \[ \ln|v + \sqrt{1 + v^2}| = \ln|x| + C \] where C is the constant of integration. We can write \(C = \ln|A|\) for some constant A. \[ \ln|v + \sqrt{1 + v^2}| = \ln|Ax| \] \[ v + \sqrt{1 + v^2} = Ax \] Substitute back \(v = \frac{y}{x}\): \[ \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = Ax \] \[ \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{|x|} = Ax \] Assuming \(x>0\) (since the initial condition is at x=1), we have: \[ \frac{y + \sqrt{x^2 + y^2}}{x} = Ax \] \[ y + \sqrt{x^2 + y^2} = Ax^2 \] Now, apply the initial condition \(y(1)=0\): \[ 0 + \sqrt{1^2 + 0^2} = A(1)^2 \] \[ 1 = A \] The particular solution is: \[ y + \sqrt{x^2 + y^2} = x^2 \] We need to find \(y(3)\). Substitute \(x=3\): \[ y(3) + \sqrt{3^2 + (y(3))^2} = 3^2 \] Let \(y(3) = y\): \[ y + \sqrt{9 + y^2} = 9 \] \[ \sqrt{9 + y^2} = 9 - y \] Square both sides: \[ 9 + y^2 = (9 - y)^2 = 81 - 18y + y^2 \] \[ 9 = 81 - 18y \] \[ 18y = 81 - 9 = 72 \] \[ y = \frac{72}{18} = 4 \] 
Step 4: Final Answer: 
The value of y(3) is 4.