Q.Solve the following differential equation: $(x + y) dy + (x - y) dx = 0; y = 1$ when $x = 1$

Concept: Homogeneous Differential Equation (degree 1 in $x$ and $y$).

Step 1 – Rewrite and check homogeneity

The equation is

The right-hand side is a function of $y/x$ only — homogeneous.

Step 2 – Substitute $y = vx$

Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting:

Step 3 – Separate variables

So

Step 4 – Integrate

This gives

Replace $v = y/x$: …

Why the homogeneous approach works

When you see a differential equation where every term in $dx$ and $dy$ has the same total degree — here, both $(x+y)$ and $(x-y)$ are degree 1 — you're looking at a homogeneous equation. The key insight: if you divide numerator and denominator by $x$ (or $y$), the equation becomes a function of the ratio $y/x$ alone. That means we can set $y = vx$, turning the problem into a separable one in $v$ and $x$.

This is powerful because it reduces a two-variable mess into a single-variable integration.

Step-by-step solution

1. Rewrite the equation in standard form

We have:

Bring the $dx$ term to the other side:

So:

This confirms homogeneity: the right-hand side is a function of $y/x$ only.

2. Substitute $y = vx$

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

Substitute into the equation:

3. Separate variables

Bring $v$ to the right:

Combine the terms on the right:

So:

Now separate:

4. Integrate both sides

Left side:

The first integral is $\tan^{-1} v$. For the second, let $u = 1 + v^2$, so $du = 2v \, dv$, giving $\frac{1}{2} \log|1 + v^2|$.

Thus:

5. Back-substitute $v = y/x$

Recall $v = y/x$, so $1 + v^2 = 1 + \frac{y^2}{x^2} = \frac{x^2 + y^2}{x^2}$.

Then:

Simplify the log term:

Plugging back:

$\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2} \log(x^2 + y^2) - \log|x| = -\log|x| + C$ …

Method: Homogeneous initial-value problem ($\arctan$ + log)

Use this for $(x+y)\,dy+(x-y)\,dx=0$ with a condition; $y=vx$ gives an inverse-tangent plus logarithm, then the point fixes the constant.

Steps

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

Rearrange to $\dfrac{dy}{dx}=\dfrac{y-x}{x+y}$, then use $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.

Step 2: Separate and split the numerator

You reach $\dfrac{1+v}{1+v^2}\,dv=-\dfrac{dx}{x}$; split into $\dfrac{1}{1+v^2}$ (→ $\tan^{-1}v$) and $\dfrac{v}{1+v^2}$ (→ $\frac12\log(1+v^2)$). …

Common Mistakes

Mistake 1: Not splitting $\dfrac{1+v}{1+v^2}$

Why it's wrong: it must break into $\frac{1}{1+v^2}$ (→ $\tan^{-1}v$) and $\frac{v}{1+v^2}$ (→ $\frac12\log(1+v^2)$). Correct approach: integrate the two pieces separately.

Mistake 2: Forgetting to apply the condition $y=1$ at $x=1$

Why it's wrong: the question wants a particular solution, so $C$ must be evaluated as $\log2+\frac{\pi}{2}$ (after doubling). Correct approach: substitute the point and simplify. …