Q.Solve the differential equation $\left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\right) \frac{dx}{dy} = 1\ (x \neq 0)$.

The key idea is to rewrite the equation in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ and solve using an integrating factor.

Step 1: Invert the given equation.

Since $\frac{dx}{dy} = 1 \big/ \frac{dy}{dx}$, we have

Step 2: Rearrange into linear form.

Here $P(x) = \frac{1}{\sqrt{x}}$ and $Q(x) = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$.

Step 3: Find the integrating factor.

Step 4: Multiply through and integrate. …

Since the bracket times $\dfrac{dx}{dy}$ equals $1$, take reciprocals to make $x$ the independent variable:

Rearrange into linear form:

Integrating factor:

Multiplying through, the left side is an exact derivative and the right side simplifies: …

Method: Invert $\dfrac{dx}{dy}$ to expose a linear equation

When the equation is written with $\dfrac{dx}{dy}$ but is really linear in $y(x)$, take reciprocals first.

Steps

Step 1: Take reciprocals.

Since $\dfrac{dy}{dx}=\dfrac{1}{\,dx/dy\,}$, rewrite the equation with $\dfrac{dy}{dx}$.

Step 2: Arrange into standard linear form.

Collect to $\dfrac{dy}{dx}+P(x)\,y=Q(x)$; here $P=\dfrac{1}{\sqrt{x}}$.

Step 3: Integrating factor. …

Common Mistakes

Mistake 1: Not inverting $\dfrac{dx}{dy}$ to reach linear form.

Why it's wrong: as written the equation hides a linear equation in $y(x)$; taking reciprocals gives $\dfrac{dy}{dx}+\tfrac{1}{\sqrt{x}}y=\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}$. Correct approach: use $\dfrac{dy}{dx}=1/\tfrac{dx}{dy}$.

Mistake 2: Getting the I.F. wrong from $P=\tfrac{1}{\sqrt{x}}$.

Why it's wrong: $\int x^{-1/2}\,dx=2\sqrt{x}$, so I.F. $=e^{2\sqrt{x}}$; a missing factor $2$ breaks the cancellation. Correct approach: integrate the power correctly. …