Home / Mathematics / Chapter 9 — Differential Equations
Class 12 Maths · Chapter 9 · Bihar Board

Differential Equations

Master differential equations with visual diagrams, order & degree, variable separable, homogeneous, linear DE methods, step-marked solutions, MCQs, A-R, and PYQs.

6 Visual Diagrams 3 Solving Methods Step Marking
5Core Concepts
20MCQs
10Assertion-Reason
8PYQs
Section 01

Basic Concepts

What is a differential equation? Order vs Degree?

9.1 Introduction to Differential Equations

An equation involving derivatives of the dependent variable with respect to the independent variable is called a differential equation.

Definition — Differential Equation

An equation involving derivatives of the dependent variable (like $y$) with respect to the independent variable (like $x$) is called a differential equation.

Example: $\frac{dy}{dx} = e^x$,   $\frac{d^2y}{dx^2} + y = 0$,   $x\frac{dy}{dx} + y = 0$

Ordinary DE: Involves derivatives with respect to only one independent variable.

Partial DE: Involves derivatives with respect to more than one independent variable.

9.2 Order of a Differential Equation

Order

The order of a differential equation is the order of the highest order derivative appearing in it.

Examples:

  • $\frac{dy}{dx} = e^x$ → Order 1
  • $\frac{d^2y}{dx^2} + y = 0$ → Order 2
  • $\left(\frac{d^3y}{dx^3}\right) + x^2\left(\frac{d^2y}{dx^2}\right)^3 = 0$ → Order 3

9.3 Degree of a Differential Equation

Degree

The degree of a differential equation (when it is a polynomial in derivatives) is the highest power (positive integer) of the highest order derivative appearing in it.

Important Points
  • The equation must be a polynomial in derivatives to define degree
  • If derivatives appear inside $\sin$, $\cos$, $\log$, $e^x$ etc., the degree is not defined
  • Order and degree (if defined) are always positive integers

Examples:

  • $\left(\frac{dy}{dx}\right)^2 + y = 0$ → Order 1, Degree 2
  • $\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$ → Order 1, Degree NOT defined

9.4 General & Particular Solutions

General Solution

The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution (or primitive).

Particular Solution

The solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution.

Key Rule
  • Order $n$ DE → General solution has $n$ arbitrary constants
  • Order 1 DE → 1 arbitrary constant
  • Order 2 DE → 2 arbitrary constants

9.5 Formation of Differential Equations

To form a DE from a given family of curves:

  1. Differentiate the given equation as many times as the number of arbitrary constants
  2. Eliminate the arbitrary constants from the equations

Example: $y = mx$ → $\frac{dy}{dx} = m \Rightarrow x\frac{dy}{dx} = y$ (eliminating $m$)

Section 02

🎨 Visual Diagrams

See differential equations in action

V1 Order vs Degree — Visual Comparison

Order = highest derivative | Degree = power of that derivative
(d²y/dx²)3 + (dy/dx)2 + y = 0
Order = 2 (highest derivative) Degree = 3 (power of highest derivative) dy/dx + sin(dy/dx) = 0 Order = 1 Degree NOT defined (derivative inside sin) Order is always a positive integer. Degree exists only if DE is polynomial in derivatives.
Order = highest derivative (always a positive integer). Degree = power of that derivative, but only if the equation is a polynomial in derivatives.

V2 Family of Solution Curves

General Solution → Family of Curves X Y C = 3 C = 2 C = 1 C = 0 Each value of C gives a different curve in the family
The general solution $y = x^2 + C$ represents an infinite family of parallel curves. Fixing C gives one particular curve.

V3 Variable Separable Method

y-terms → dy side | x-terms → dx side dy/dx = g(x) · h(y) y-terms dy / h(y) x-terms g(x) dx = ↓ Integrate ∫ dy/h(y) = ∫ g(x) dx + C
The variable separable method: separate the variables to opposite sides of the equal sign, then integrate both sides.

V4 Homogeneous DE — Substitution $y = vx$

Convert to variable separable Homogeneous DE dy/dx = F(x,y) → y = vx After substitution v + x·dv/dx = g(v) ↓ Variable Separable dv / [g(v)-v] = dx/x Substitute back v = y/x to get the final solution
Homogeneous DEs are solved by substituting $y = vx$, which turns them into variable separable form.

V5 Linear DE — Integrating Factor

Multiply by I.F. to make LHS exact dy/dx + P·y = Q Step 1: Integrating Factor I.F. = e^(∫P dx) ↓ Step 2: Multiply by I.F. d/dx [y · I.F.] = Q · I.F. y · (I.F.) = ∫ (Q · I.F.) dx + C
The Integrating Factor converts the LHS of a linear DE into a perfect derivative, making it integrable in one step.

V6 Real World Applications of DEs

Where are differential equations used? ⚛️ Physics Motion, Waves Heat flow 🧬 Biology Population growth 💰 Economics Interest rates Investment models 🏗️ Engineering Circuits Structures 💊 Medicine Drug decay Epidemiology ⚗️ Chemistry Reaction rates Concentration
Differential equations model real-world phenomena across physics, biology, economics, engineering, medicine, and chemistry.
Section 03

Methods of Solving

Three main methods for first-order, first-degree DEs

M1 Variable Separable Method

If $\frac{dy}{dx} = g(x) \cdot h(y)$, separate the variables:

$$\frac{dy}{h(y)} = g(x) \, dx$$

Then integrate both sides:

$$\int \frac{dy}{h(y)} = \int g(x) \, dx + C$$
When to use
  • When you can move all y-terms to one side and x-terms to the other
  • Function on RHS can be factored as $g(x) \cdot h(y)$

M2 Homogeneous Differential Equations

A DE of the form $\frac{dy}{dx} = F(x, y)$ is homogeneous if $F(x, y)$ is a homogeneous function of degree zero.

Test: Replace $x \to \lambda x$, $y \to \lambda y$. If $F(\lambda x, \lambda y) = F(x, y)$, the DE is homogeneous.

Substitution: Let $y = vx$ (or $x = vy$ if easier). Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

This reduces the DE to variable separable form in $v$ and $x$.

Standard Form

$\frac{dy}{dx} = g\left(\frac{y}{x}\right)$ — a function of $\frac{y}{x}$ only

M3 Linear Differential Equations

A DE of the form:

$$\frac{dy}{dx} + P y = Q$$

where P and Q are functions of $x$ only (or constants).

Integrating Factor (I.F.):

$$\text{I.F.} = e^{\int P \, dx}$$

Solution:

$$y \cdot (\text{I.F.}) = \int (Q \cdot \text{I.F.}) \, dx + C$$
Alternate Form

If $\frac{dx}{dy} + P_1 x = Q_1$ (P₁, Q₁ functions of $y$):

I.F. $= e^{\int P_1 dy}$ and $x \cdot (\text{I.F.}) = \int (Q_1 \cdot \text{I.F.}) dy + C$

Quick Reference

📐 All Important Formulas

Everything you need from Chapter 9

F1 Basic Concepts

Order = Highest order derivative in DE
Degree = Power of highest order derivative (if polynomial)
General solution has $n$ arbitrary constants (for order $n$)
Particular solution has 0 arbitrary constants

F2 Variable Separable Method

$$\frac{dy}{dx} = g(x) \cdot h(y)$$
$$\int \frac{dy}{h(y)} = \int g(x) \, dx + C$$

F3 Homogeneous DE

$$\frac{dy}{dx} = g\left(\frac{y}{x}\right)$$
Substitute: $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$
$$\frac{dv}{g(v) - v} = \frac{dx}{x}$$

F4 Linear DE

$$\frac{dy}{dx} + Py = Q$$
$$\text{I.F.} = e^{\int P \, dx}$$
$$y \cdot (\text{I.F.}) = \int (Q \cdot \text{I.F.}) \, dx + C$$

F5 Standard Integrals (for DEs)

$\int \frac{dx}{x} = \log|x| + C$
$\int e^{ax} dx = \frac{e^{ax}}{a} + C$
$\int \frac{dx}{1 + x^2} = \tan^{-1}x + C$
$\int \sec^2 x \, dx = \tan x + C$

F6 Applications

Population growth: $\frac{dP}{dt} = kP \Rightarrow P = P_0 e^{kt}$
Radioactive decay: $\frac{dN}{dt} = -kN$
Newton's cooling: $\frac{dT}{dt} = -k(T - T_a)$
Shortcuts

💡 Tricks & Shortcuts

Speed techniques for board exams

Trick 1 — Identify Order & Degree

"Highest derivative = Order, its power = Degree"

  • Scan for the highest derivative first
  • Then check the power of that derivative
  • If any derivative is inside $\sin$, $\cos$, $\log$, $e$ → Degree is NOT defined
Trick 2 — Which Method to Use?

Decision tree:

  • Can you separate $x$ and $y$? → Variable Separable
  • Is RHS a function of $y/x$ only? → Homogeneous
  • Is it in form $\frac{dy}{dx} + Py = Q$? → Linear
Trick 3 — Homogeneous Test

"Every term must have same total degree"

  • Replace $x \to \lambda x$ and $y \to \lambda y$
  • If you can factor out $\lambda^n$ (with same $n$ for all terms) → Homogeneous
  • Otherwise → Not homogeneous
Trick 4 — I.F. Memory

"P for the y-coefficient after writing in standard form"

  • Always write DE as $\frac{dy}{dx} + Py = Q$ first
  • P = coefficient of $y$ (with sign!)
  • I.F. $= e^{\int P dx}$
Trick 5 — Number of Constants

"Order = Number of arbitrary constants in general solution"

  • Order 1 → 1 constant
  • Order 2 → 2 constants
  • Particular solution → 0 constants
Trick 6 — Quick Variable Separable

Look for multiplicative structure

  • If RHS looks like $f(x) \times g(y)$, separate directly
  • Common patterns: $\frac{dy}{dx} = e^{x+y}$, $\frac{dy}{dx} = xy$
  • Rewrite $e^{x+y} = e^x \cdot e^y$
Trick 7 — Integration Shortcut for DEs

Common integrals you'll need:

  • $\int \frac{dy}{y} = \log|y| + C$
  • $\int y^n dy = \frac{y^{n+1}}{n+1} + C$ (for $n \neq -1$)
  • $\int \frac{dy}{1+y^2} = \tan^{-1}y + C$
Trick 8 — Order & Degree Trap

Watch out for these common mistakes:

  • $\sin\left(\frac{dy}{dx}\right)$ → degree NOT defined (not polynomial in derivative)
  • $e^{dy/dx}$ → degree NOT defined
  • $\left(\frac{d^2y}{dx^2}\right)^{1/2}$ → degree NOT defined (not integer power)
Trick 9 — Substitution Choice

For homogeneous DEs:

  • Use $y = vx$ when $x$ appears more "conveniently"
  • Use $x = vy$ when $y$ appears more "conveniently"
  • Check which substitution makes the resulting DE simpler
Trick 10 — Boundary Conditions

To find the particular solution:

  • First find the general solution (with constant $C$)
  • Substitute the given point $(x_0, y_0)$ into it
  • Solve for $C$
  • Replace $C$ in the general solution
Practice

📝 Solved Examples with Step Marking

Click "Show Answer" to reveal step-by-step solutions with marks

📋 Total: 2 Marks · Board Pattern

Q1. Find the order and degree of $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + y = 0$.

Step 1 — Find the highest derivative0.75 Mark

The highest order derivative present is $\frac{d^2y}{dx^2}$ (second derivative).

So the Order = 2.

Step 2 — Find the degree0.75 Mark

The equation is a polynomial in derivatives (all powers are non-negative integers).

The power of the highest order derivative $\frac{d^2y}{dx^2}$ is $1$.

So the Degree = 1.

Order = 2, Degree = 1
📋 Total: 3 Marks

Q2. Solve the differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$.

Step 1 — Separate variables1 Mark

Since $1 + y^2 \neq 0$, we can write:

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

Step 2 — Integrate both sides1 Mark

$\int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2}$

$\tan^{-1}y = \tan^{-1}x + C$

Step 3 — Write final solution1 Mark

$\tan^{-1}y = \tan^{-1}x + C$ is the general solution.

$\tan^{-1}y - \tan^{-1}x = C$
📋 Total: 3 Marks

Q3. Find the general solution of $\frac{dy}{dx} - y = \cos x$.

Step 1 — Identify P and Q0.5 Mark

The DE is of the form $\frac{dy}{dx} + Py = Q$ with $P = -1$, $Q = \cos x$.

Step 2 — Find I.F.0.75 Mark

I.F. $= e^{\int P \, dx} = e^{\int -1 \, dx} = e^{-x}$

Step 3 — Multiply and simplify0.75 Mark

$e^{-x}\frac{dy}{dx} - e^{-x}y = e^{-x}\cos x$

$\frac{d}{dx}(y \cdot e^{-x}) = e^{-x}\cos x$

Step 4 — Integrate1 Mark

$y \cdot e^{-x} = \int e^{-x}\cos x \, dx + C$

Using $\int e^{ax}\cos bx \, dx = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2 + b^2}$:

$y e^{-x} = \frac{e^{-x}(\sin x - \cos x)}{2} + C$

$y = \frac{\sin x - \cos x}{2} + Ce^{x}$

$y = \frac{\sin x - \cos x}{2} + Ce^x$
Test Yourself

✍️ MCQ Practice — 20 Questions

Select your answer and click "Check Answers"

1. The order of $\frac{d^2y}{dx^2} + y = 0$ is:

Correct: B — Highest derivative is $\frac{d^2y}{dx^2}$, so order = 2.

2. The degree of $\left(\frac{dy}{dx}\right)^3 + y = 0$ is:

Correct: C — Power of the highest derivative is 3.

3. The degree of $\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$ is:

Correct: A — Since the derivative appears inside sin, it's not a polynomial in derivatives.

4. The general solution of $\frac{dy}{dx} = 2x$ is:

Correct: B — Integrate: $\int 2x\,dx = x^2 + C$.

5. A DE with order 3 has how many arbitrary constants in its general solution?

Correct: C — Number of arbitrary constants = order of the DE.

6. In linear DE $\frac{dy}{dx} + Py = Q$, the Integrating Factor is:

Correct: A — Standard formula: I.F. = $e^{\int P\,dx}$.

7. $\frac{dy}{dx} = \frac{x+y}{x-y}$ is:

Correct: D — RHS is a function of $y/x$ only, so it's homogeneous.

8. In the homogeneous substitution $y = vx$, $\frac{dy}{dx}$ equals:

Correct: B — Differentiate $y = vx$ using product rule.

9. The DE $\frac{dy}{dx} = xy$ is:

Correct: C — Can be written as $\frac{dy}{y} = x\,dx$, variables separated.

10. The solution of $\frac{dy}{dx} = e^{x+y}$ is:

Correct: A — $e^{-y}dy = e^x dx$ → $-e^{-y} = e^x + C_1$ → $e^x + e^{-y} = C$.

11. $\frac{d^3y}{dx^3} + \left(\frac{dy}{dx}\right)^4 = 0$ has order and degree:

Correct: D — Highest derivative is $d^3y/dx^3$ (order 3), its power is 1 (degree 1).

12. A solution that contains arbitrary constants is called:

Correct: B — General solution contains arbitrary constants.

13. For the DE $\frac{dy}{dx} + y \cot x = 2x$ , P equals:

Correct: A — In standard form $\frac{dy}{dx} + Py = Q$, P = $\cot x$.

14. The order of $(y'')^2 + (y')^3 + y^4 = 0$ is:

Correct: C — $y''$ is the highest derivative (order 2).

15. $\frac{dy}{dx} = \frac{y}{x}$ is:

Correct: B — RHS is a function of $y/x$ only, so it's homogeneous (and variable separable too).

16. General solution of $\frac{dy}{dx} + y = 1$ is:

Correct: D — Linear DE: I.F. $= e^x$. Solution: $ye^x = e^x + C$ → $y = 1 + Ce^{-x}$.

17. Solution of $\frac{dy}{dx} = y$ is:

Correct: A — $\frac{dy}{y} = dx \Rightarrow \log y = x + C \Rightarrow y = Ce^x$.

18. Population growth model $\frac{dP}{dt} = kP$ has solution:

Correct: B — Exponential growth: $P = P_0 e^{kt}$ or $P = Ce^{kt}$.

19. If $y = A\cos x + B\sin x$, the DE is:

Correct: C — Differentiating twice gives $y'' = -A\cos x - B\sin x = -y$, so $y'' + y = 0$.

20. Number of arbitrary constants in the particular solution of a DE of order 2 is:

Correct: A — Particular solution has 0 arbitrary constants (they are fixed by initial conditions).
Score: 0 / 20
New Pattern

🎯 Assertion & Reasoning — 10 Questions

Each question has an Assertion (A) and a Reason (R). Choose the correct option.

📖 Standard Options

(A) Both A and R are true, and R is the correct explanation of A.

(B) Both A and R are true, but R is NOT the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

1
Assertion (A) The order of $\frac{d^2y}{dx^2} + y = 0$ is 2.
Reason (R) Order is the order of the highest order derivative in the differential equation.
Correct: A — Both true and R is the definition that explains A.
2
Assertion (A) The degree of $\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0$ is 1.
Reason (R) The degree of a DE is defined only when it is a polynomial in derivatives.
Correct: D — A is false (degree not defined because of sin). R is true (this is the correct rule).
3
Assertion (A) The general solution of a first-order DE contains exactly 1 arbitrary constant.
Reason (R) The number of arbitrary constants in the general solution equals the order of the DE.
Correct: A — Both true, R is the underlying rule for A.
4
Assertion (A) $\frac{dy}{dx} = \frac{x+y}{x-y}$ is a homogeneous differential equation.
Reason (R) The RHS can be written as a function of $y/x$ only.
Correct: A — Both true. $\frac{x+y}{x-y} = \frac{1 + y/x}{1 - y/x}$ which is a function of $y/x$.
5
Assertion (A) In a linear DE $\frac{dy}{dx} + Py = Q$, the I.F. is $e^{\int P dx}$.
Reason (R) Multiplying by this I.F. makes the LHS the derivative of $y \cdot \text{I.F.}$
Correct: A — Both true. R explains why the I.F. formula works.
6
Assertion (A) The DE $\frac{dy}{dx} + y = x$ is homogeneous.
Reason (R) A homogeneous DE has the form $\frac{dy}{dx} = g(y/x)$.
Correct: D — A is false (it's a linear DE, not homogeneous). R is true (defining condition of homogeneous DE).
7
Assertion (A) The solution of $\frac{dy}{dx} = \frac{1}{y}$ is $\frac{y^2}{2} = x + C$.
Reason (R) This DE is variable separable.
Correct: A — Both true. $y\,dy = dx$ gives $\frac{y^2}{2} = x + C$.
8
Assertion (A) The number of arbitrary constants in the particular solution is 0.
Reason (R) Particular solutions are obtained by assigning specific values to the arbitrary constants.
Correct: A — Both true. R explains why A holds.
9
Assertion (A) The DE $\frac{dy}{dx} = xy$ is variable separable.
Reason (R) It can be rewritten as $\frac{dy}{y} = x\,dx$.
Correct: A — Both true. The rewriting proves variable separability.
10
Assertion (A) In the homogeneous DE substitution, $y = vx$, we get $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Reason (R) This follows from differentiating $y = vx$ using the product rule.
Correct: A — Both true, R is the derivation of A.
Score: 0 / 10
Exam Focus

📖 Previous Year Questions

Frequently asked questions in board exams

2024

Q. Find the order and degree of $\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + y = 0$.

Step 1 — Highest derivative0.5 Mark

Highest derivative = $\frac{d^2y}{dx^2}$ → Order = 2

Step 2 — Degree0.5 Mark

Power of $\frac{d^2y}{dx^2}$ = 3 → Degree = 3

Order = 2, Degree = 3
2024

Q. Solve $\frac{dy}{dx} = \frac{x+1}{2-y}$, $(y \neq 2)$.

Step 1 — Separate variables1 Mark

$(2-y)dy = (x+1)dx$

Step 2 — Integrate1.5 Marks

$\int(2-y)dy = \int(x+1)dx$

$2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C_1$

Step 3 — Rearrange0.5 Mark

Multiply by 2: $4y - y^2 = x^2 + 2x + C$

Or: $x^2 + y^2 + 2x - 4y + C = 0$

$x^2 + y^2 + 2x - 4y + C = 0$
2023

Q. Show that the DE $(x-y)\frac{dy}{dx} = x + 2y$ is homogeneous and solve it.

Step 1 — Rewrite and test homogeneity1 Mark

$\frac{dy}{dx} = \frac{x+2y}{x-y}$. Replacing $x \to \lambda x$, $y \to \lambda y$ gives the same form → homogeneous.

Step 2 — Substitute y = vx1 Mark

$v + x\frac{dv}{dx} = \frac{1+2v}{1-v}$

$x\frac{dv}{dx} = \frac{1+2v}{1-v} - v = \frac{v^2+v+1}{1-v}$

$\frac{1-v}{v^2+v+1} dv = \frac{dx}{x}$

Step 3 — Integrate1.5 Marks

$\int \frac{1-v}{v^2+v+1} dv = \int \frac{dx}{x}$

Completing the square and integrating gives the general solution in terms of $\log$ and $\tan^{-1}$.

$\log|(x^2+xy+y^2)| = 2\sqrt{3}\tan^{-1}\left(\frac{x+2y}{\sqrt{3}x}\right) + C$
2023

Q. Solve the linear DE $\frac{dy}{dx} - y = \cos x$.

Step 1 — Identify P, Q0.5 Mark

P = $-1$, Q = $\cos x$

Step 2 — Find I.F.0.5 Mark

I.F. $= e^{\int -1 dx} = e^{-x}$

Step 3 — Multiply and integrate1.5 Marks

$\frac{d}{dx}(ye^{-x}) = e^{-x}\cos x$

$ye^{-x} = \frac{e^{-x}(\sin x - \cos x)}{2} + C$

$y = \frac{\sin x - \cos x}{2} + Ce^{x}$
2022

Q. Solve $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.

Step 1 — Separate variables1 Mark

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

Step 2 — Integrate1 Mark

$\tan^{-1}y = \tan^{-1}x + C$

$\tan^{-1}y - \tan^{-1}x = C$
2022

Q. Find the general solution of the DE $\frac{dy}{dx} + 2y = \sin x$.

Step 1 — P and Q0.5 Mark

P = 2, Q = $\sin x$

Step 2 — I.F.0.5 Mark

I.F. $= e^{\int 2 dx} = e^{2x}$

Step 3 — Solve2 Marks

$ye^{2x} = \int e^{2x}\sin x\,dx + C$

$= \frac{e^{2x}(2\sin x - \cos x)}{5} + C$

$y = \frac{2\sin x - \cos x}{5} + Ce^{-2x}$
2021

Q. Verify that $y = e^{-3x}$ is a solution of $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.

Step 1 — Find derivatives1 Mark

$y = e^{-3x}$, $\frac{dy}{dx} = -3e^{-3x}$, $\frac{d^2y}{dx^2} = 9e^{-3x}$

Step 2 — Substitute in DE1 Mark

LHS $= 9e^{-3x} + (-3e^{-3x}) - 6e^{-3x} = 9e^{-3x} - 9e^{-3x} = 0$ = RHS ✓

Verified — $y = e^{-3x}$ is a solution
2020

Q. Find the particular solution of $\frac{dy}{dx} = -4xy^2$, given that $y = 1$ when $x = 0$.

Step 1 — Separate variables1 Mark

$\frac{dy}{y^2} = -4x\,dx$

Step 2 — Integrate1 Mark

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

$\frac{1}{y} = 2x^2 + C'$ (where $C' = -C$)

Step 3 — Apply condition1 Mark

At $x = 0, y = 1$: $1 = 0 + C' \Rightarrow C' = 1$

$y = \frac{1}{2x^2+1}$

$y = \frac{1}{2x^2+1}$