Solve the following differential equation: $\frac{dy}{dx} = (1+x^2)(1+y^2)$

> [!TLDR] > This is a **separable** first‑order ODE. By separating variables and integrating both sides, we obtain $\tan^{-1} y = x + \frac{x^3}{3} + C$, so the general solution is $y = \tan\left(x + \frac{x^3}{3} + C\right)$. The key insight: the right‑hand side is a product of a function of $x$ alone and a function of $y$ alone. That’s the hallmark of a **separable** differential equation. Instead of trying to solve it as a whole, we can “split” the derivative into two separate integrals — one over $y$, one over $x$ — and then recombine. Let’s walk through it. --- 1. **Rewrite the equation in differential form** The derivative $\frac{dy}{dx}$ is a ratio. Multiply both sides by $dx$ to get $$dy = (1+x^2)(1+y^2)\,dx.$$ Now divide both sides by $(1+y^2)$ (which is never zero for real $y$, so no special cases are lost): $$\frac{dy}{1+y^2} = (1+x^2)\,dx.$$ The variables are now separated — all $y$’s on the left, all $x$’s on the right. 2. **Integrate both sides** The left side is a standard integral: $$\int \frac{dy}{1+y^2} = \tan^{-1} y + C_1.$$ The right side is a simple polynomial: $$\int (1+x^2)\,dx = x + \frac{x^3}{3} + C_2.$$ Combine the two constants into a single constant $C = C_2 - C_1$: $$\tan^{-1} y = x + \frac{x^3}{3} + C.$$ 3. **Solve for $y$ explicitly** Take the tangent of both sides. Since $\tan(\tan^{-1} y) = y$, we get $$y = \tan\left(x + \frac{x^3}{3} + C\right).$$ …

Solve the following differential equation: (dy)/(dx) = (1+x²)(1+y²)

Exercise 9.3 · Q6

\frac{dy}{dx} = (1+x^2)(1+y^2)

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Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

Imagine you're baking a cake. The recipe says "mix the dry ingredients separately, then add the wet ones." You keep things that belong together together, and things that don't apart — until the right moment. Separation of Variables does exactly that for certain kinds of equations.

The Core Intuition

Some equations involve two different kinds of change happening at once. Think of a cup of hot coffee cooling down. The rate at which it cools depends on:

The temperature difference between the coffee and the room (a function of time)
The surface area of the cup (a function of shape, not time)

These two influences are tangled together in one equation. Separation of Variables is the mathematical trick that untangles them — it lets you handle the time part first, then the space part separately.

The Precise Statement

Separation of Variables applies to ordinary differential equations (ODEs) of the form:

\frac{dy}{dx} = f(x) \cdot g(y)

where the right-hand side is a product of a function of $x$ alone and a function of $y$ alone. The method works in three clean steps:

\frac{dy}{dx} = f(x) \cdot g(y) \quad \Longrightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx

Step 1: Separate. Multiply both sides by $dx$ and divide by $g(y)$ (assuming $g(y) \neq 0$ ). This moves all $y$ 's to one side and all $x$ 's to the other.

Step 2: Integrate. Put an integral sign on both sides:

\int \frac{1}{g(y)} \, dy = \int f(x) \, dx

Step 3: Solve. Evaluate both integrals and solve for $y$ explicitly if possible.

Watch out

You cannot separate if the equation is not in product form. For example, $\frac{dy}{dx} = x + y$ cannot be separated — the sum $x+y$ is not a product $f(x)g(y)$ .

Why This Works

The justification is the chain rule in reverse. From $\frac{dy}{dx} = f(x)g(y)$ , rewrite it as:

\frac{1}{g(y)} \frac{dy}{dx} = f(x)

Now integrate both sides with respect to $x$ :

\int \frac{1}{g(y)} \frac{dy}{dx} \, dx = \int f(x) \, dx

The left side is a substitution waiting to happen: $\frac{dy}{dx} dx = dy$ , so you get $\int \frac{1}{g(y)} dy$ . That's the entire trick — the chain rule dressed up.

A Concrete Example

Solve $\frac{dy}{dx} = 2xy$ , with $y(0) = 3$ .

Step 1: Separate. Divide both sides by $y$ (assuming $y \neq 0$ ):

\frac{1}{y} \, dy = 2x \, dx

Step 2: Integrate.

\int \frac{1}{y} \, dy = \int 2x \, dx \quad\Longrightarrow\quad \log|y| = x^2 + C

Step 3: Solve for $y$ .

|y| = e^{x^2 + C} = e^C \cdot e^{x^2}

Let $A = \pm e^C$ (absorbing the absolute value): $y = A e^{x^2}$ . Now use $y(0) = 3$ : $3 = A e^{0} = A$ , so $A = 3$ .

Final answer: $\boxed{y = 3e^{x^2}}$ …

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This is a separable first‑order ODE. By separating variables and integrating both sides, we obtain $\tan^{-1} y = x + \frac{x^3}{3} + C$ , so the general solution is $y = \tan\left(x + \frac{x^3}{3} + C\right)$ .

The key insight: the right‑hand side is a product of a function of $x$ alone and a function of $y$ alone. That’s the hallmark of a separable differential equation. Instead of trying to solve it as a whole, we can “split” the derivative into two separate integrals — one over $y$ , one over $x$ — and then recombine.

Let’s walk through it.

Rewrite the equation in differential form The derivative $\frac{dy}{dx}$ is a ratio. Multiply both sides by $dx$ to get

dy = (1+x^2)(1+y^2)\,dx.

Now divide both sides by $(1+y^2)$ (which is never zero for real $y$ , so no special cases are lost):

\frac{dy}{1+y^2} = (1+x^2)\,dx.

The variables are now separated — all $y$ ’s on the left, all $x$ ’s on the right.

Integrate both sides The left side is a standard integral:

\int \frac{dy}{1+y^2} = \tan^{-1} y + C_1.

The right side is a simple polynomial:

\int (1+x^2)\,dx = x + \frac{x^3}{3} + C_2.

Combine the two constants into a single constant $C = C_2 - C_1$ :

\tan^{-1} y = x + \frac{x^3}{3} + C.

Solve for $y$ explicitly Take the tangent of both sides. Since $\tan(\tan^{-1} y) = y$ , we get $y = \tan\left(x + \frac{x^3}{3} + C\right).$ …

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

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

Method: Separation into two products (variables separable)

Steps

Common Mistakes

Q.Solve the following differential equation: dydx=(1+x2)(1+y2)\frac{dy}{dx} = (1+x^2)(1+y^2)dxdy​=(1+x2)(1+y2)

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

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