Solve the differential equation $\frac{dy}{dx}+2xy=y$.

Solve the differential equation (dy)/(dx)+2xy=y.

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\frac{dy}{dx}+2xy=y

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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 first-order linear ODE solved by separation of variables. The general solution is $y = Ce^{x - x^2}$ , where $C$ is an arbitrary constant.

The equation $\frac{dy}{dx} + 2xy = y$ looks like it might need an integrating factor — but before jumping into that, notice something: the $y$ on the right-hand side can be brought over to the left. That gives us a chance to factor $y$ out entirely. When you can write the derivative in terms of $y$ times something, separation of variables is often the cleanest path.

Let’s rewrite it:

\frac{dy}{dx} + 2xy - y = 0 \quad \Rightarrow \quad \frac{dy}{dx} + y(2x - 1) = 0.

Now it’s clear: the derivative of $y$ plus $y$ times a function of $x$ equals zero. That’s a separable equation.

Separate the variables. Move the $y$ term to the other side:

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

Divide both sides by $y$ (assuming $y \neq 0$ for now; we’ll check the zero case later) and multiply by $dx$ :

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

Integrate both sides. The left integrates to $\log|y|$ , the right is a simple polynomial:

\int \frac{1}{y}\, dy = \int -(2x - 1)\, dx.

\log|y| = -x^2 + x + C_1,

where $C_1$ is the constant of integration.

Solve for $y$ . Exponentiate both sides to remove the logarithm:

|y| = e^{-x^2 + x + C_1} = e^{C_1} e^{x - x^2}.

Let $C = \pm e^{C_1}$ (or simply an arbitrary constant, since $e^{C_1} > 0$ and the $\pm$ absorbs the absolute value). Then:

y = C e^{x - x^2}.

Check the special case $y = 0$ . …

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Q.Solve the differential equation $\frac{dy}{dx}+2xy=y$ .

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 a first-order equation by factoring the derivative

Steps

Common Mistakes

Q.Solve the differential equation dydx+2xy=y\frac{dy}{dx}+2xy=ydxdy​+2xy=y.

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Q.Solve the differential equation $\frac{dy}{dx}+2xy=y$ .

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