Solve initial value problem using given boundary condition.

Find the Particular Solution given y(0) = 1

Question

Solve the differential equation:

\frac{dy}{dx} = 2xy

given the initial condition $y(0) = 1$ . Find the particular solution.

This type — a separable ODE with an initial value — appears regularly in CBSE 12 board exams and as warm-up questions in JEE Main. The 2025 CBSE Sample Paper carried a nearly identical question for 3 marks.

Solution — Step by Step

We want all $y$ terms on one side and all $x$ terms on the other. Divide both sides by $y$ (valid as long as $y \neq 0$ ):

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

This is the core move in variable separable ODEs — we’re splitting the equation so each side can be integrated independently.

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

\ln|y| = x^2 + C

where $C$ is the constant of integration. We write one constant on the right — it absorbs constants from both sides.

Raise $e$ to both sides:

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

Let $A = \pm e^C$ (which is just an arbitrary non-zero constant). So the general solution is:

y = A\,e^{x^2}

Now we use what we know: when $x = 0$ , $y = 1$ .

1 = A\,e^{0} = A \cdot 1 \implies A = 1

Substituting $A = 1$ :

\boxed{y = e^{x^2}}

This is the particular solution satisfying $y(0) = 1$ .

Why This Works

The equation $\frac{dy}{dx} = 2xy$ says: the rate of change of $y$ is proportional to both $y$ itself and to $x$ . As $x$ grows, the function grows faster and faster — which is exactly the behaviour of $e^{x^2}$ .

The general solution $y = Ae^{x^2}$ is a family of curves, one for each value of $A$ . The initial condition $y(0) = 1$ pins down which specific curve passes through the point $(0, 1)$ . This is the geometric meaning of an initial value problem.

Every particular solution is a member of this family. Change the initial condition — say $y(0) = 3$ — and you’d get $y = 3e^{x^2}$ , the same shape but scaled vertically.

Alternative Method — Using the Integrating Factor

While variable separation is the natural approach here, we can also treat this as a linear ODE of the form $\frac{dy}{dx} + P(x)\,y = Q(x)$ .

Rewrite: $\frac{dy}{dx} - 2x\,y = 0$ , so $P(x) = -2x$ , $Q(x) = 0$ .

Integrating factor: $\mu = e^{\int -2x\,dx} = e^{-x^2}$

Multiply through: $\frac{d}{dx}\left(y \cdot e^{-x^2}\right) = 0$

Integrate: $y \cdot e^{-x^2} = C \implies y = Ce^{x^2}$

Applying $y(0) = 1$ gives $C = 1$ , so $y = e^{x^2}$ .

For linear ODEs where $Q(x) = 0$ , the integrating factor method and variable separation always give the same result. In exams, use whichever you can execute faster — for CBSE, variable separation saves 30–40 seconds.

Common Mistake

Students often write $\ln y = x^2 + C$ and then directly apply the initial condition as $\ln(1) = 0 + C$ , getting $C = 0$ , and conclude $y = e^{x^2}$ . The answer is correct here — but only by luck. The right process is to first exponentiate to get $y = Ae^{x^2}$ , then substitute. If the initial condition were $y(0) = 3$ , the shortcut gives $C = \ln 3$ , so $y = e^{x^2 + \ln 3} = 3e^{x^2}$ — messy but correct. Getting into the habit of solving for $A$ first keeps errors out of your board answer sheet.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Integrating Factor

Common Mistake

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