Q.Solve the following differential equation: $y = x \sin x$ ; $xy' = y + x \sqrt{x^2 - y^2}$ ($x \neq 0$ and $x > y$ or $x < -y$)

The key idea is that the equation is homogeneous — it can be written in the form $y' = f(y/x)$.

Step 1: Rewrite the equation.

Given $xy' = y + x\sqrt{x^2 - y^2}$, divide through by $x$ (since $x \neq 0$):

Step 2: Express in terms of $v = y/x$.

Let $y = vx$, so $y' = v + x v'$. Substituting:

Since $x > y$ or $x < -y$, we have $|v| < 1$ and $|x| = x$ for $x>0$ (or $|x| = -x$ for $x<0$). For simplicity, take $x > 0$; the sign can be absorbed later. Then:

Step 3: Integrate.

Integrating both sides:

Thus $v = \sin(x + C)$, and since $v = y/x$: …

Why the substitution works

Putting $y=vx$ makes $\sqrt{x^2-y^2}=|x|\sqrt{1-v^2}$, and the $\frac{y}{x}=v$ term cancels, leaving a separable equation in $v$ and $x$.

Step-by-step solution

1. Substitute $y=vx$, so $y' = v + x\dfrac{dv}{dx}$. The equation becomes

2. Simplify. Since $\sqrt{x^2-v^2x^2}=|x|\sqrt{1-v^2}$,

Dividing by $x^2$ (with $\tfrac{|x|}{x}=1$ for $x>0$):

3. Integrate. …

Method: Solve a homogeneous equation by the substitution $y=vx$

Use this when $\frac{dy}{dx}$ can be written as a function of $\frac{y}{x}$ alone, e.g. $xy'=y+x\sqrt{x^2-y^2}$.

Steps

Step 1: Put $y=vx$, so $\frac{dy}{dx}=v+x\frac{dv}{dx}$.

Substitute; the plain $\frac{y}{x}=v$ part cancels against the same term on the right, leaving only a $v$-and-$x$ relation.

Step 2: Separate and integrate. …

Common Mistakes

Mistake 1: Trying to separate before substituting $y=vx$.

Why it's wrong: $xy'=y+x\sqrt{x^2-y^2}$ is homogeneous, not directly separable; only after $y=vx$ does the $\frac{y}{x}$ part cancel and leave a separable $v$-equation. Correct approach: substitute $y=vx$, $y'=v+xv'$ first.

Mistake 2: Mishandling $\sqrt{x^2-y^2}$ under the substitution. …