Question

Let the solution curve of the differential equation \[ x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx,\quad x>0, \] with $y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to

• A. 4
• B. 2
• C. 1
• D. 6

Hint:

Differential equations of the form $x\,dy - y\,dx = f(\sqrt{x^2+y^2})\,dx$ are best solved using the substitution $y=vx$.

Answer 1

The Correct Option is A

The given differential equation is \[ x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx \] Step 1: Rewrite the equation in standard form.
Dividing both sides by $dx$, we get \[ x\frac{dy}{dx} - y = \sqrt{x^2+y^2} \] Step 2: Use the substitution $y = vx$.
Let \[ y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \] Substituting into the equation: \[ x(v + x\frac{dv}{dx}) - vx = \sqrt{x^2 + v^2x^2} \] \[ x^2\frac{dv}{dx} = x\sqrt{1+v^2} \] Step 3: Separate the variables.
\[ \frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x} \] Step 4: Integrate both sides.
\[ \int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x} \] \[ \sinh^{-1}(v) = \ln x + C \] Step 5: Apply the initial condition.
Given $y(1)=0$, \[ v = \frac{y}{x} = 0 \text{ at } x=1 \] \[ \sinh^{-1}(0) = \ln 1 + C \Rightarrow C=0 \] Thus, \[ \sinh^{-1}(v)=\ln x \] Step 6: Express $y$ in terms of $x$.
\[ v = \sinh(\ln x)=\frac{x-\frac{1}{x}}{2} \] \[ y = vx = \frac{x^2-1}{2} \] Step 7: Evaluate $y(3)$.
\[ y(3)=\frac{9-1}{2}=4 \] Final Answer: $\boxed{4}$