(v) General solution of the differential equation of the type $\frac{dx}{dy}+P_1 x=Q_1$ is given by ______.

> [!TLDR] > The general solution of a first-order linear differential equation in $x$ is found using an integrating factor. The solution is $x \cdot (\text{I.F.}) = \int Q_1 \cdot (\text{I.F.}) \, dy + C$, where the integrating factor $\text{I.F.} = e^{\int P_1 \, dy}$. This question tests a standard result from differential equations. The key is recognising the form: when the derivative is $\frac{dx}{dy}$ (rather than $\frac{dy}{dx}$), the independent variable becomes $y$, and the dependent variable is $x$. The equation $\frac{dx}{dy} + P_1 x = Q_1$ is a **first-order linear differential equation** in $x$. Here, $P_1$ and $Q_1$ are functions of $y$ alone (or constants). The method of solution is identical in spirit to the more familiar form $\frac{dy}{dx} + P y = Q$ — we just swap the roles of $x$ and $y$. 1. **The core idea**: We multiply the entire equation by a specially chosen function called the **integrating factor (I.F.)**. This factor makes the left-hand side a perfect derivative (of a product), which we can then integrate directly. 2. **Finding the integrating factor**: For the standard form $\frac{dx}{dy} + P_1 x = Q_1$, the integrating factor is: $$ \text{I.F.} = e^{\int P_1 \, dy} $$ Why? Because if we multiply the equation by $e^{\int P_1 \, dy}$, the left side becomes the derivative of $x \cdot e^{\int P_1 \, dy}$ with respect to $y$. Let's verify: $$ \frac{d}{dy} \left( x \cdot e^{\int P_1 \, dy} \right) = \frac{dx}{dy} \cdot e^{\int P_1 \, dy} + x \cdot P_1 e^{\int P_1 \, dy} = e^{\int P_1 \, dy} \left( \frac{dx}{dy} + P_1 x \right) $$ This matches the left-hand side of the original equation after multiplication. 3. **Applying the method**: Multiply the given equation $\frac{dx}{dy} + P_1 x = Q_1$ by the integrating factor: $$ e^{\int P_1 \, dy} \cdot \frac{dx}{dy} + P_1 x e^{\int P_1 \, dy} = Q_1 e^{\int P_1 \, dy} $$ The left side simplifies as shown above: $$ \frac{d}{dy} \left( x \cdot e^{\int P_1 \, dy} \right) = Q_1 e

(v) General solution of the differential equation of the type (dx)/(dy)+P₁ x=Q₁ is given by ___.

NCERT Exemplar · Q38

\frac{dx}{dy}+P_1 x=Q_1

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Concept understanding — Integrating Factor Method

Integrating Factor Method

Some first-order differential equations refuse to separate — you cannot get all the $y$ 's on one side and all the $x$ 's on the other. The integrating factor method is the standard trick for a special (and very common) family of these: the linear first-order equation. The idea is beautifully simple: multiply the whole equation by one cleverly chosen function, and the messy left-hand side collapses into a single derivative that we can integrate directly.

The Standard Form

An equation is linear of first order if it can be written as

\frac{dy}{dx} + P\,y = Q

where $P$ and $Q$ are functions of $x$ alone (or constants). Notice $y$ and $\frac{dy}{dx}$ appear only to the first power, and never multiplied together — that is what "linear" means here.

Tip

Always rearrange into this exact shape first. The coefficient of $\frac{dy}{dx}$ must be $1$ before you read off $P$ and $Q$ .

The Integrating Factor

The magic multiplier is

\text{I.F.} = e^{\int P\,dx}.

Why this one? Multiply the equation by $e^{\int P\,dx}$ :

e^{\int P\,dx}\frac{dy}{dx} + P\,e^{\int P\,dx}\,y = Q\,e^{\int P\,dx}.

By the product rule, the left-hand side is exactly $\dfrac{d}{dx}\!\left(y\cdot e^{\int P\,dx}\right)$ , because the derivative of $e^{\int P\,dx}$ is $P\,e^{\int P\,dx}$ . So the equation becomes

\frac{d}{dx}\big(y\cdot \text{I.F.}\big) = Q\cdot \text{I.F.}

The left side is now a single derivative — that is the whole point of choosing this factor.

The Solution

Integrate both sides with respect to $x$ :

y\cdot \text{I.F.} = \int \big(Q\cdot \text{I.F.}\big)\,dx + C.

This is the general solution. In words: (solution) × (integrating factor) = integral of (Q × integrating factor), plus a constant.

A Quick Illustration

For $\dfrac{dy}{dx} + \dfrac{1}{x}\,y = x$ , we read $P = \tfrac1x$ , $Q = x$ . Then $\int P\,dx = \log x$ , so $\text{I.F.} = e^{\log x} = x$ . The solution is

y\cdot x = \int x\cdot x\,dx + C = \frac{x^3}{3} + C.

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The general solution of a first-order linear differential equation in $x$ is found using an integrating factor. The solution is $x \cdot (\text{I.F.}) = \int Q_1 \cdot (\text{I.F.}) \, dy + C$ , where the integrating factor $\text{I.F.} = e^{\int P_1 \, dy}$ .

This question tests a standard result from differential equations. The key is recognising the form: when the derivative is $\frac{dx}{dy}$ (rather than $\frac{dy}{dx}$ ), the independent variable becomes $y$ , and the dependent variable is $x$ .

The equation $\frac{dx}{dy} + P_1 x = Q_1$ is a first-order linear differential equation in $x$ . Here, $P_1$ and $Q_1$ are functions of $y$ alone (or constants). The method of solution is identical in spirit to the more familiar form $\frac{dy}{dx} + P y = Q$ — we just swap the roles of $x$ and $y$ .

The core idea: We multiply the entire equation by a specially chosen function called the integrating factor (I.F.). This factor makes the left-hand side a perfect derivative (of a product), which we can then integrate directly.
Finding the integrating factor: For the standard form $\frac{dx}{dy} + P_1 x = Q_1$ , the integrating factor is:

\text{I.F.} = e^{\int P_1 \, dy}

Why? Because if we multiply the equation by $e^{\int P_1 \, dy}$ , the left side becomes the derivative of $x \cdot e^{\int P_1 \, dy}$ with respect to $y$ . Let's verify:

\frac{d}{dy} \left( x \cdot e^{\int P_1 \, dy} \right) = \frac{dx}{dy} \cdot e^{\int P_1 \, dy} + x \cdot P_1 e^{\int P_1 \, dy} = e^{\int P_1 \, dy} \left( \frac{dx}{dy} + P_1 x \right)

This matches the left-hand side of the original equation after multiplication.

Applying the method: Multiply the given equation $\frac{dx}{dy} + P_1 x = Q_1$ by the integrating factor:

e^{\int P_1 \, dy} \cdot \frac{dx}{dy} + P_1 x e^{\int P_1 \, dy} = Q_1 e^{\int P_1 \, dy}

The left side simplifies as shown above:

\frac{d}{dy} \left( x \cdot e^{\int P_1 \, dy} \right) = Q_1 e^{\int P_1 \, dy}

Integrating both sides with respect to $y$ : $\int \frac{d}{dy} \left( x \cdot e^{\int P_1 \, dy} \right) \, dy = \int Q_1 e^{\int P_1 \, dy} \, dy$ …

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Q.(v) General solution of the differential equation of the type $\frac{dx}{dy}+P_1 x=Q_1$ is given by ______.

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Solving a First-Order Equation That Is Linear in $x$

Steps

Common Mistakes

Q.(v) General solution of the differential equation of the type dxdy+P1x=Q1\frac{dx}{dy}+P_1 x=Q_1dydx​+P1​x=Q1​ is given by ______.

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Q.(v) General solution of the differential equation of the type $\frac{dx}{dy}+P_1 x=Q_1$ is given by ______.

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