Q.Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $y = Ax$ : $xy' = y \ (x \neq 0)$

The key idea is that a homogeneous linear differential equation of the form $xy' = y$ has solutions of the form $y = Cx$.

Step 1: Differentiate the given function $y = Ax$ with respect to $x$:

Step 2: Substitute $y'$ and $y$ into the left-hand side of the differential equation:

$xy' = x \cdot A = Ax$ …

We are asked to verify that $y = Ax$ (where $A$ is an arbitrary constant) is a solution of the differential equation $xy' = y$, with $x \neq 0$.

This is a verification problem, not a solving problem. The given function is already proposed as a solution; we just need to check that it satisfies the differential equation. The equation $xy' = y$ is a first-order ordinary differential equation. It is also a homogeneous differential equation (in the sense that it can be written as $y' = y/x$, which is a function of $y/x$ alone), but here we are simply substituting.

The core idea: a solution to a differential equation is any function that, when plugged in along with its derivatives, makes the equation true for all $x$ in the domain. So we take the candidate $y = Ax$, compute its derivative $y'$, substitute both into $xy' = y$, and see if the equality holds identically.

Let’s go step by step.

1. Write down the candidate function.

   We have $y = Ax$, where $A$ is a constant (real number). This represents a family of straight lines through the origin, each with slope $A$.

2. Differentiate with respect to $x$.

   Since $A$ is constant,

The derivative is simply the constant $A$.

3. Substitute into the left-hand side of the differential equation. The left-hand side is $xy'$. Replace $y'$ with $A$:

4. Compare with the right-hand side. The right-hand side is $y$, which is $Ax$ (from the candidate). So we have:

5. Check the domain condition. …

Method: Verify a straight-line solution of $xy'=y$

Use this for candidates like $y=Ax$ against $xy'=y$.

Steps

Step 1: Differentiate the candidate.

$y=Ax\Rightarrow y'=A$ (a constant, since $A$ is a fixed parameter).

Step 2: Substitute into each side of the equation.

Left: $xy'=x\cdot A=Ax$. Right: $y=Ax$.

Step 3: Compare over the stated domain. …

Common Mistakes

Mistake 1: Differentiating $A$ as if it varied.

Why it's wrong: in $y=Ax$, $A$ is a fixed parameter, so $y'=A$, a constant. Treating $A$ as a function of $x$ corrupts the check. Correct approach: $\frac{d}{dx}(Ax)=A$.

Mistake 2: Ignoring the domain restriction $x\neq 0$. …