Solve the differential equation $\frac{dy}{dx}=1+x+y^2+xy^2$, when $y=0$, $x=0$.

> [!TLDR] > The given equation is separable after factoring the right-hand side as $(1+x)(1+y^2)$. Integrating both sides and applying the initial condition gives $y = \tan\left(x + \frac{x^2}{2}\right)$. The key insight here is **separation of variables**. Many first-order differential equations that look messy at first glance can be rewritten so that all terms involving $y$ (and $dy$) are on one side and all terms involving $x$ (and $dx$) are on the other. Once separated, we integrate each side independently. Let’s see how this works for $$\frac{dy}{dx} = 1 + x + y^2 + xy^2.$$ 1. **Factor the right-hand side** Group the terms cleverly: $$1 + x + y^2 + xy^2 = (1 + x) + y^2(1 + x) = (1 + x)(1 + y^2).$$ So the equation becomes $$\frac{dy}{dx} = (1 + x)(1 + y^2).$$ 2. **Separate the variables** Multiply both sides by $dx$ and divide by $(1 + y^2)$ (which is never zero for real $y$, so no division issues): $$\frac{dy}{1 + y^2} = (1 + x)\,dx.$$ Now the variables are isolated — $y$ on the left, $x$ on the right. 3. **Integrate both sides** The left integral is a standard form: $$\int \frac{dy}{1 + y^2} = \tan^{-1} y + C_1.$$ The right side is straightforward: $$\int (1 + x)\,dx = x + \frac{x^2}{2} + C_2.$$ Combining constants into a single constant $C$, we get $$\tan^{-1} y = x + \frac{x^2}{2} + C.$$ 4. **Apply the initial condition** We are given $y = 0$ when $x = 0$. Substitute: $$\tan^{-1}(0) = 0 + 0 + C \quad\Rightarrow\quad 0 = C.$$ So $C = 0$. 5. **Solve for $y$** With $C = 0$, we have $$\tan^{-1} y = x + \frac{x^2}{2}.$$ …

Solve the differential equation (dy)/(dx)=1+x+y²+xy², when y=0, x=0.

NCERT Exemplar · Q9

\frac{dy}{dx}=1+x+y^2+xy^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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The given equation is separable after factoring the right-hand side as $(1+x)(1+y^2)$ . Integrating both sides and applying the initial condition gives $y = \tan\left(x + \frac{x^2}{2}\right)$ .

The key insight here is separation of variables. Many first-order differential equations that look messy at first glance can be rewritten so that all terms involving $y$ (and $dy$ ) are on one side and all terms involving $x$ (and $dx$ ) are on the other. Once separated, we integrate each side independently.

Let’s see how this works for

\frac{dy}{dx} = 1 + x + y^2 + xy^2.

Factor the right-hand side Group the terms cleverly:

1 + x + y^2 + xy^2 = (1 + x) + y^2(1 + x) = (1 + x)(1 + y^2).

So the equation becomes

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

Separate the variables Multiply both sides by $dx$ and divide by $(1 + y^2)$ (which is never zero for real $y$ , so no division issues):

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

Now the variables are isolated — $y$ on the left, $x$ on the right.

Integrate both sides The left integral is a standard form:

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

The right side is straightforward:

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

Combining constants into a single constant $C$ , we get

\tan^{-1} y = x + \frac{x^2}{2} + C.

Apply the initial condition We are given $y = 0$ when $x = 0$ . Substitute:

\tan^{-1}(0) = 0 + 0 + C \quad\Rightarrow\quad 0 = C.

So $C = 0$ .

Solve for $y$ With $C = 0$ , we have $\tan^{-1} y = x + \frac{x^2}{2}.$ …

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Q.Solve the differential equation $\frac{dy}{dx}=1+x+y^2+xy^2$ , when $y=0$ , $x=0$ .

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: Separating by factoring a four-term right side, with a condition

Steps

Common Mistakes

Q.Solve the differential equation dydx=1+x+y2+xy2\frac{dy}{dx}=1+x+y^2+xy^2dxdy​=1+x+y2+xy2, when y=0y=0y=0, x=0x=0x=0.

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Q.Solve the differential equation $\frac{dy}{dx}=1+x+y^2+xy^2$ , when $y=0$ , $x=0$ .

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