Solve the following differential equation:

$(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}; \ y=0 ext{ when } x=1$

Question

Solve the following differential equation:

$(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}; \ y=0 	ext{ when } x=1$

Accepted Answer

> [!TLDR]
> This is a first-order linear ODE solved by rewriting it in standard form, finding the integrating factor $\mu(x)=1+x^2$, and then integrating both sides. The particular solution satisfying $y(1)=0$ is $y = \frac{	an^{-1}x - \frac{\pi}{4}}{1+x^2}$.

The problem gives us a differential equation with an initial condition — an **Initial Value Problem (IVP)**. The equation is:

$$(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}, \quad y(1)=0$$

When you see a first-order ODE where the left side looks like it might be the derivative of a product, your first instinct should be: **can I write this as $\frac{d}{dx}[	ext{something}]$?** That’s the heart of the integrating factor method.

Let’s check: the derivative of $y \cdot (1+x^2)$ is $(1+x^2)\frac{dy}{dx} + 2xy$ — exactly the left-hand side! So the equation is already in “exact derivative” form. No need to hunt for an integrating factor; it’s already there.

1. **Rewrite as a perfect derivative**  
   Notice that:
   $$\frac{d}{dx}\left[y(1+x^2)ight] = (1+x^2)\frac{dy}{dx} + 2xy$$
   So the ODE becomes:
   $$\frac{d}{dx}\left[y(1+x^2)ight] = \frac{1}{1+x^2}$$

2. **Integrate both sides with respect to $x$**  
   $$\int \frac{d}{dx}\left[y(1+x^2)ight]\,dx = \int \frac{1}{1+x^2}\,dx$$
   The left side gives $y(1+x^2)$. The right side is a standard integral:
   $$\int \frac{1}{1+x^2}\,dx = 	an^{-1}x + C$$
   So:
   $$y(1+x^2) = 	an^{-1}x + C$$

3. **Apply the initial condition $y(1)=0$**  
   Substitute $x=1$, $y=0$:
   $$0 \cdot (1+1^2) = 	an^{-1}(1) + C \quad\Rightarrow\quad 0 = \frac{\pi}{4} + C$$
   Hence $C = -\frac{\pi}{4}$.

4. **Write the particular solution**  
   $$y(1+x^2) = 	an^{-1}x - \frac{\pi}{4}$$
   Therefore: …

Q.Solve the following differential equation: $(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}; \ y=0 \text{ when } x=1$

Concept understanding — Initial Value Problem

Initial Value Problem

The intuition: a rule of change plus a starting point

The precise statement

A worked example

Method: Exact-derivative linear IVP, then apply the condition

Steps

Common Mistakes

Q.Solve the following differential equation: (1+x2)dydx+2xy=11+x2; y=0 when x=1(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}; \ y=0 \text{ when } x=1(1+x2)dxdy​+2xy=1+x21​; y=0 when x=1

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Q.Solve the following differential equation: $(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}; \ y=0 \text{ when } x=1$

Unlock everything free for 14 days