Q.Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.

Concept: Differential Equation Modeling — translating a geometric condition into a differential equation and solving it with an initial condition.

The slope of the tangent at $(x, y)$ is $\frac{dy}{dx}$. The condition says:

This is a first-order linear differential equation. Rewrite as:

The integrating factor is $e^{\int -1\,dx} = e^{-x}$. Multiply through:

Integrate both sides: …

Why this approach works

The slope of the tangent to a curve at any point $(x, y)$ is simply $\frac{dy}{dx}$. The problem tells us this slope equals the sum of the coordinates, i.e., $x + y$. So we have:

This is a differential equation — an equation linking a function, its derivative, and the independent variable. Solving it means finding the function $y(x)$ that satisfies this relationship and also passes through the origin $(0,0)$.

The equation $\frac{dy}{dx} = x + y$ is a first-order linear differential equation. It's not separable in its current form (you can't get all $y$ terms on one side and all $x$ terms on the other), so we need a systematic method: the integrating factor method.

Step-by-step solution

1. Write the equation in standard form

We have $\frac{dy}{dx} = x + y$. Bring the $y$ term to the left:

Here $P(x) = -1$ and $Q(x) = x$.

2. Find the integrating factor

3. Multiply both sides by $\mu(x)$

The left side is now the derivative of $y \cdot e^{-x}$ (check by differentiating: $\frac{d}{dx}(y e^{-x}) = e^{-x}\frac{dy}{dx} - y e^{-x}$). So:

4. Integrate both sides

The right side requires integration by parts. Let $u = x$, $dv = e^{-x}dx$. Then $du = dx$, $v = -e^{-x}$.

So:

$y e^{-x} = -x e^{-x} - e^{-x} + C$ …

Method: Turning a geometric slope condition into a differential equation

For "the slope at $(x,y)$ equals ..." problems, translate the words into a differential equation, solve it, and use the stated point as the initial condition.

Steps

Step 1: Replace 'slope of the tangent' with $\dfrac{dy}{dx}$.

Write the stated relationship as an equation in $x,y,\dfrac{dy}{dx}$.

Step 2: Identify the type. …

Common Mistakes

Mistake 1: Misreading "sum of the coordinates" as a product or a single coordinate.

Why it's wrong: the slope equals $x+y$, so $\dfrac{dy}{dx}=x+y$; using $xy$ or just $x$ gives the wrong equation. Correct approach: translate "sum of coordinates" as $x+y$.

Mistake 2: Treating $\dfrac{dy}{dx}=x+y$ as separable. …