Solve the following differential equation: $(x+y)\frac{dy}{dx}=1$

Solve the following differential equation: (x+y)(dy)/(dx)=1

Exercise 9.5 · Q10

(x+y)\frac{dy}{dx}=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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Treat $x$ as a function of $y$ : the equation becomes linear, giving $x=Ce^{y}-y-1$ .

Spotting the trick

Written as $\frac{dy}{dx}=\frac{1}{x+y}$ the equation is neither separable nor linear in $y$ . But $\frac{dx}{dy}=\frac{1}{dy/dx}$ , so flipping the derivative turns it linear in $x$ .

Set up

(x+y)\frac{dy}{dx}=1 \implies \frac{dx}{dy} = x + y \implies \frac{dx}{dy} - x = y.

This is $\frac{dx}{dy}+P(y)x=Q(y)$ with $P=-1$ , $Q=y$ .

Integrating factor

\mu(y) = e^{\int(-1)\,dy} = e^{-y}.

Multiply and integrate

\frac{d}{dy}\big(x\,e^{-y}\big) = y\,e^{-y}.

By parts ( $u=y,\ dv=e^{-y}dy$ ),

\int y\,e^{-y}\,dy = -y e^{-y} - e^{-y} + C.

So $x\,e^{-y} = -y e^{-y} - e^{-y} + C$ . …

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Q.Solve the following differential equation: $(x+y)\frac{dy}{dx}=1$

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Spotting the trick

Set up

Integrating factor

Multiply and integrate

Method: Flip to $\dfrac{dx}{dy}$ when the equation is linear in $x$

Steps

Common Mistakes

Q.Solve the following differential equation: (x+y)dydx=1(x+y)\frac{dy}{dx}=1(x+y)dxdy​=1

Spotting the trick

Set up

Integrating factor

Multiply and integrate

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Q.Solve the following differential equation: $(x+y)\frac{dy}{dx}=1$

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