Q.(ii) Solution of the differential equation of the type $\frac{dx}{dy}+p_1 x=Q_1$ is given by $x\cdot(\text{I.F.})=\int (\text{I.F.})\times Q_1\,dy$. (State True or False.)

The key idea is that a first-order linear differential equation can be written in the form $\frac{dx}{dy} + p_1(y) x = Q_1(y)$, and its solution follows the standard integrating factor method.

Step 1: For $\frac{dx}{dy} + p_1 x = Q_1$, the integrating factor (I.F.) is $e^{\int p_1 \, dy}$.

Step 2: The general solution is obtained by multiplying both sides by the I.F. and integrating: …

The core idea here is recognising the type of differential equation. When you see $\frac{dx}{dy}$ with a term involving $x$ alone (like $p_1 x$) on the left, and a function of $y$ alone ($Q_1$) on the right, you're looking at a first-order linear differential equation — but with $x$ as the dependent variable and $y$ as the independent variable.

Most textbooks first teach the form $\frac{dy}{dx} + P y = Q$, where $y$ is a function of $x$. But the roles of variables can be swapped. The structure is identical: the derivative of the dependent variable appears linearly, the dependent variable itself appears linearly (multiplied by a function of the independent variable), and the right-hand side is a function of the independent variable only.

The method of integrating factor works because of the product rule in reverse. For $\frac{dx}{dy} + p_1 x = Q_1$, we multiply both sides by $\text{I.F.} = e^{\int p_1 \, dy}$. The left side then becomes $\frac{d}{dy}(x \cdot \text{I.F.})$, which integrates directly. That's why the formula $x \cdot (\text{I.F.}) = \int (\text{I.F.}) \times Q_1 \, dy$ holds.

Let's verify step by step.

1. Identify the form. The given equation is $\frac{dx}{dy} + p_1 x = Q_1$, where $p_1$ and $Q_1$ are functions of $y$ (or constants). This is a linear differential equation of first order in $x$.

2. Recall the standard solution for $\frac{dy}{dx} + P y = Q$. For that form, the integrating factor is $\text{I.F.} = e^{\int P \, dx}$, and the solution is $y \cdot (\text{I.F.}) = \int (\text{I.F.}) \times Q \, dx$. This is a proven result.

3. Swap variables. If we replace $y$ with $x$ and $x$ with $y$, the form $\frac{dx}{dy} + p_1 x = Q_1$ is exactly analogous. The integrating factor becomes $\text{I.F.} = e^{\int p_1 \, dy}$, and the solution formula becomes $x \cdot (\text{I.F.}) = \int (\text{I.F.}) \times Q_1 \, dy$.

4. Why the formula works. Multiply the equation by $\text{I.F.}$:

   $\text{I.F.} \cdot \frac{dx}{dy} + p_1 \cdot \text{I.F.} \cdot x = \text{I.F.} \cdot Q_1$ …

Method: Writing the General Solution of the $\frac{dx}{dy}$-Linear Form

Once the integrating factor of a linear-in-$x$ equation is known, the general solution follows a fixed template: (variable) × (I.F.) = integral of (Q × I.F.).

Steps

Step 1: Start from the standard form and its integrating factor.

For $\frac{dx}{dy}+p_1x=Q_1$, the integrating factor is $\text{I.F.}=e^{\int p_1\,dy}$.

Step 2: Collapse the left side to a single derivative.

Multiplying through gives

$\frac{d}{dy}\big(x\cdot\text{I.F.}\big)=Q_1\cdot\text{I.F.}$ …