Q.Solve the following differential equation: $x dy - y dx = \sqrt{x^2 + y^2} dx$

The key idea is that the equation is homogeneous — it can be written in the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$.

Step 1: Rewrite the equation.

Divide through by $dx$ (assuming $x \neq 0$):

Step 2: Substitute $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

The equation becomes:

Step 3: Simplify and separate variables:

$x\frac{dv}{dx} = \sqrt{1+v^2} \quad\Rightarrow\quad \frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}.$ …

1. Why this is a homogeneous equation

A first-order differential equation is called homogeneous if it can be written in the form

$\frac{dy}{dx} = F\left(\frac{y}{x}\right)$.

Here, the given equation is:

Divide through by $dx$:

So

The right-hand side depends only on $y/x$ because:

That’s exactly $F(y/x)$. So the substitution $y = vx$ will work.

2. Substitute $y = vx$

Let $y = vx$, where $v$ is a function of $x$. Then:

Plug into the equation:

Simplify the RHS:

Cancel $v$ from both sides:

3. Separate variables

Now we have a separable equation:

Integrate both sides.

We’ll use the logarithmic form because it’s more convenient later.

4. Integrate

Left side:

Right side:

$\int \frac{dx}{x} = \log|x| + \log C$ …

Method: Homogeneous equation with a square root ($\sqrt{x^2+y^2}$)

Use this for $x\,dy-y\,dx=\sqrt{x^2+y^2}\,dx$, where the root is homogeneous of degree 1, so $y=vx$ reduces it to a $\dfrac{dv}{\sqrt{1+v^2}}$ integral.

Steps

Step 1: Solve for $\dfrac{dy}{dx}$ and substitute $y=vx$

Rearrange to $\dfrac{dy}{dx}=\dfrac{y+\sqrt{x^2+y^2}}{x}$, then substitute; the root becomes $x\sqrt{1+v^2}$ (for $x>0$).

Step 2: Cancel and separate

The $v$ cancels, leaving $x\dfrac{dv}{dx}=\sqrt{1+v^2}$, i.e. $\dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{x}$. …

Common Mistakes

Mistake 1: Pulling $x$ out of $\sqrt{x^2+y^2}$ as $x$ instead of $|x|$

Why it's wrong: $\sqrt{x^2+y^2}=|x|\sqrt{1+v^2}$; the sign of $x$ matters for the domain. Correct approach: state the branch (e.g. $x>0$) so $\sqrt{x^2+y^2}=x\sqrt{1+v^2}$.

Mistake 2: Misremembering $\int\dfrac{dv}{\sqrt{1+v^2}}$ …