Q.Solve the following differential equation: $y' = \frac{x+y}{x}$

The key idea is that this is a first-order Initial Value Problem (though no initial condition is given here, so we find the general solution). The equation is not separable in its current form, but it can be rewritten as a linear ODE.

Step 1: Rewrite the equation.

$y' = \frac{x+y}{x} = 1 + \frac{y}{x}$

So $y' - \frac{1}{x} y = 1$.

Step 2: Identify the integrating factor.

Here $P(x) = -\frac{1}{x}$, so the integrating factor is

$\mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$.

Step 3: Multiply through and integrate.

$\frac{1}{x} y' - \frac{1}{x^2} y = \frac{1}{x}$

The left side is $\frac{d}{dx}\left( \frac{y}{x} \right)$.

Integrate: $\frac{y}{x} = \int \frac{1}{x} dx = \log|x| + C$.

Step 4: Solve for $y$.

$y = x \log|x| + Cx$.

The key here is to recognize that the right-hand side $\frac{x+y}{x}$ can be split into two simpler terms: $1 + \frac{y}{x}$. That immediately reveals the equation is not separable in its current form, but it is linear in $y$.

An Initial Value Problem (IVP) isn’t given here — we’re just solving the differential equation generally. But the approach for a first-order linear ODE is always the same: rewrite it as $y' + P(x)y = Q(x)$, then multiply through by an integrating factor $\mu(x) = e^{\int P(x)\,dx}$ to make the left side a perfect derivative.

Let’s walk through it.

1. Rewrite the equation in standard linear form. Start with $y' = \frac{x+y}{x} = 1 + \frac{y}{x}$. Bring the $y$ term to the left:

So here $P(x) = -\frac{1}{x}$ and $Q(x) = 1$.

2. Find the integrating factor. Compute $\int P(x)\,dx = \int -\frac{1}{x}\,dx = -\log|x| = \log|x|^{-1}$. Then the integrating factor is:

For simplicity, we usually take $\mu(x) = \frac{1}{x}$ (assuming $x>0$; the absolute value can be handled later with a sign).

3. Multiply the entire equation by $\mu(x)$.

Notice the left side is exactly the derivative of $\frac{y}{x}$:

So the equation becomes:

4. Integrate both sides.

where $C$ is the constant of integration.

5. Solve for $y$. Multiply through by $x$:

Method: $y=vx$ for a homogeneous (initial-value) equation

Use this for equations like $y'=\dfrac{x+y}{x}=1+\dfrac{y}{x}$ that depend only on $\dfrac{y}{x}$; the $y=vx$ substitution reduces them to a simple separable equation.

Steps

Step 1: Substitute $y=vx$

With $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ and the right side $=1+v$, the equation becomes $v+x\dfrac{dv}{dx}=1+v$.

Step 2: Cancel and separate

The $v$ terms cancel, leaving $x\dfrac{dv}{dx}=1$, i.e. $dv=\dfrac{dx}{x}$.

Step 3: Integrate and restore $y$

Integrate to $v=\log|x|+C$, then $y=vx=x\log|x|+Cx$. Apply any initial condition to fix $C$.

Common Mistakes

Mistake 1: Missing the cancellation of $v$

Why it's wrong: after substituting, both sides carry $v$; cancelling gives the simple $x\frac{dv}{dx}=1$. Failing to cancel over-complicates it. Correct approach: subtract $v$ from both sides.

Mistake 2: Forgetting the modulus in $\log|x|$

Why it's wrong: the integral of $\frac{1}{x}$ is $\log|x|$; dropping the modulus loses part of the domain. Correct approach: write $v=\log|x|+C$.

Mistake 3: Not restoring $y=vx$

Why it's wrong: the solution must be in $x,y$: $y=x\log|x|+Cx$. Correct approach: multiply $v$ back by $x$.