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

We need to solve the differential equation given by

\[\ x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx,\quad x>0\]

with the initial condition \( y(1) = 0 \).

First, let's rewrite the given differential equation:

\[x \frac{dy}{dx} - y = \sqrt{x^2 + y^2}\]

We will rearrange the differential equation to make it more tractable:

\[x \frac{dy}{dx} = y + \sqrt{x^2 + y^2}\]

Divide through by \(x\):

\[\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x}\]

Consider the substitution \( y = vx \), where \( v \) is a function of \( x \). Then,

\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]

Substitute back into the equation:

\[v + x\frac{dv}{dx} = \frac{vx}{x} + \frac{\sqrt{x^2 + (vx)^2}}{x}\]

This simplifies to:

\[v + x\frac{dv}{dx} = v + \sqrt{1 + v^2}\]

Canceling the \( v \) terms, we get:

\[x\frac{dv}{dx} = \sqrt{1 + v^2}\]

Separate variables to integrate:

\[\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}\]

Integrate both sides:

- The left side becomes: \(\int \frac{dv}{\sqrt{1+v^2}} = \sinh^{-1}(v)\) - The right side becomes: \(\int \frac{dx}{x} = \ln|x| + C\)

Equating the results:

\[\sinh^{-1}(v) = \ln|x| + C\]

Since \( y = vx \), substitute \( v = \frac{y}{x} \):

- \(v = \sinh(\ln|x| + C)\)

Using the initial condition \( y(1) = 0 \) implies \( y(1) = v(1) \cdot 1 = 0 \):

- Hence, \( v(1) = 0 \) - \( \sinh^{-1}(0) = \ln|1| + C \Rightarrow C = 0 \)

Conclude that:

\[v = \sinh(\ln x)\]

Final expression for \( y \):

\[y = x \cdot \sinh(\ln x)\]

To find \( y(3) \):

- \(y(3) = 3 \cdot \sinh(\ln 3)\)

Using the identity \(\sinh(\ln x) = \frac{x^2 - 1}{2x}\), we get:

- \(y(3) = 3 \cdot \frac{3^2 - 1}{2 \cdot 3} = 3 \cdot \frac{8}{6} = 4\)

Thus, the value of \( y(3) \) is 4.