Differential Equations — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Differential Equations" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Differential Equations" — 8 important questions with detailed answers for CBSE boa…
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NCERT Solutions for Class 12 Mathematics — Differential Equations. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Differential Equations — All 8 Questions Solved
Q1. Solve the differential equation dy/dx = 2x with the initial condition y(0) = 1.
To solve dy/dx = 2x with y(0) = 1:
Step 1: Separate variables (already separated).
dy = 2x dx
Step 2: Integrate both sides.
∫ dy = ∫ 2x dx
y = x² + C
Step 3: Apply initial condition y(0) = 1.
1 = 0² + C
C = 1
Step 4: Write the particular solution.
y = x² + 1
Final Answer: y = x² + 1
Q2. Solve the differential equation dy/dx + 2y = e^(-2x).
To solve linear DE: dy/dx + 2y = e^(-2x):
Step 1: Identify the form.
This is dy/dx + P(x)y = Q(x)
where P(x) = 2 and Q(x) = e^(-2x)
Step 2: Find integrating factor.
I.F. = e^(∫ P(x) dx)
= e^(∫ 2 dx)
= e^(2x)
Step 3: Multiply the equation by I.F.
e^(2x) × dy/dx + 2e^(2x) × y = e^(2x) × e^(-2x)
e^(2x) × dy/dx + 2e^(2x) × y = 1
Step 4: Recognize left side as derivative of product.
d/dx[e^(2x) × y] = 1
Step 5: Integrate both sides.
e^(2x) × y = ∫ 1 dx
e^(2x) × y = x + C
Step 6: Solve for y.
y = (x + C)e^(-2x)
Final Answer: y = (x + C)e^(-2x)
Q3. Solve the differential equation (x + y) dx = (x + y + 1) dy.
To solve (x + y) dx = (x + y + 1) dy:
Step 1: Rearrange the equation.
(x + y) dx - (x + y + 1) dy = 0
Step 2: Use substitution.
Let u = x + y, then du = dx + dy
So dx = du - dy
Step 3: Substitute in the equation.
u(du - dy) - (u + 1) dy = 0
u du - u dy - u dy - dy = 0
u du - 2u dy - dy = 0
u du = (2u + 1) dy
Step 4: Separate variables.
u/(2u + 1) du = dy
Step 5: Integrate left side.
∫ u/(2u + 1) du = ∫ dy
For left integral, let 2u + 1 = t, then u = (t - 1)/2 and du = dt/2
= ∫ [(t - 1)/2]/t × dt/2
= 1/4 ∫ (t - 1)/t dt
= 1/4 ∫ (1 - 1/t) dt
= 1/4 [t - ln|t|] + C₁
= 1/4 [2u + 1 - ln|2u + 1|] + C₁
Step 6: Equate both sides.
1/4 [2u + 1 - ln|2u + 1|] = y + C₂
1/4 [2(x + y) + 1 - ln|2(x + y) + 1|] = y + C
2(x + y) + 1 - ln|2(x + y) + 1| = 4y + 4C
2x + 2y + 1 - ln|2x + 2y + 1| = 4y + C₃
2x - 2y + 1 - ln|2x + 2y + 1| = C₃
Final Answer: 2x - 2y + 1 = ln|2x + 2y + 1| + C
Q4. Solve dy/dx = (x + y)/(x - y).
To solve dy/dx = (x + y)/(x - y):
Step 1: Use substitution.
Let y = vx, then dy/dx = v + x(dv/dx)
Step 2: Substitute in the equation.
v + x(dv/dx) = (x + vx)/(x - vx)
v + x(dv/dx) = x(1 + v)/[x(1 - v)]
v + x(dv/dx) = (1 + v)/(1 - v)
Step 3: Rearrange.
x(dv/dx) = (1 + v)/(1 - v) - v
x(dv/dx) = [(1 + v) - v(1 - v)]/(1 - v)
x(dv/dx) = [1 + v - v + v²]/(1 - v)
x(dv/dx) = (1 + v²)/(1 - v)
Step 4: Separate variables.
(1 - v)/(1 + v²) dv = dx/x
Step 5: Integrate both sides.
∫ (1 - v)/(1 + v²) dv = ∫ dx/x
∫ 1/(1 + v²) dv - ∫ v/(1 + v²) dv = ln|x| + C₁
tan⁻¹(v) - (1/2)ln(1 + v²) = ln|x| + C₁
Step 6: Substitute back v = y/x.
tan⁻¹(y/x) - (1/2)ln(1 + y²/x²) = ln|x| + C
tan⁻¹(y/x) - (1/2)ln(x² + y²)/x²] = ln|x| + C
tan⁻¹(y/x) - (1/2)ln(x² + y²) + ln|x| = ln|x| + C
tan⁻¹(y/x) = (1/2)ln(x² + y²) + C
Final Answer: tan⁻¹(y/x) = (1/2)ln(x² + y²) + C
Q5. Solve the differential equation x(dy/dx) = y + x² cos x.
To solve x(dy/dx) = y + x² cos x:
Step 1: Rewrite in standard form.
dy/dx - y/x = x cos x
dy/dx + (-1/x)y = x cos x
Step 2: Find integrating factor.
I.F. = e^(∫ -1/x dx) = e^(-ln|x|) = 1/x
Step 3: Multiply equation by I.F.
(1/x)(dy/dx) - y/x² = cos x
Step 4: Recognize left side as derivative.
d/dx[y/x] = cos x
Step 5: Integrate both sides.
y/x = ∫ cos x dx
y/x = sin x + C
Step 6: Solve for y.
y = x sin x + Cx
Final Answer: y = x(sin x + C)
Q6. Solve the differential equation d²y/dx² - 5(dy/dx) + 6y = 0.
To solve second-order linear ODE: d²y/dx² - 5(dy/dx) + 6y = 0:
Step 1: Write the characteristic equation.
Assuming y = e^(mx), we get:
m² - 5m + 6 = 0
Step 2: Solve the characteristic equation.
Using factorization:
(m - 2)(m - 3) = 0
m = 2 or m = 3
Step 3: Write the general solution.
Since we have two distinct real roots, the general solution is:
y = C₁e^(2x) + C₂e^(3x)
where C₁ and C₂ are arbitrary constants.
Final Answer: y = C₁e^(2x) + C₂e^(3x)
Q7. Find the order and degree of the differential equation d³y/dx³ + (d²y/dx²)² - 3(dy/dx) = sin x.
To find order and degree:
Step 1: Identify the order.
The order is the highest derivative in the equation.
Highest derivative: d³y/dx³
Order = 3
Step 2: Identify the degree.
Degree is the power of the highest order derivative, after clearing fractions and removing radicals.
In the equation: d³y/dx³ + (d²y/dx²)² - 3(dy/dx) = sin x
The highest order derivative d³y/dx³ appears with power 1.
Step 3: Check if degree can be assigned.
The equation is a polynomial in derivatives, and d³y/dx³ has power 1.
Degree = 1
Final Answer: Order = 3, Degree = 1
Q8. Solve the differential equation e^y dx - (xe^y + 2y) dy = 0.
To solve e^y dx - (xe^y + 2y) dy = 0:
Step 1: Rearrange.
e^y dx = (xe^y + 2y) dy
Step 2: Check if it's exact.
Rewrite as: e^y dx - (xe^y + 2y) dy = 0
M(x, y) = e^y, N(x, y) = -(xe^y + 2y)
∂M/∂y = e^y
∂N/∂x = -e^y
∂M/∂y ≠ ∂N/∂x, so it's not exact.
Step 3: Try separating variables or finding integrating factor.
Rewrite: e^y dx - xe^y dy - 2y dy = 0
e^y(dx - x dy) = 2y dy
Step 4: Recognize the pattern.
d(xe^y) = e^y dx + x × e^y dy
So e^y(dx - x dy) = e^y dx - xe^y dy
This suggests dividing by e^(2y):
dx/e^y - x/e^y × dy = 2y/e^y dy
Alternatively, rewrite as:
e^y dx - xe^y dy = 2y dy
d(xe^y) = 2y dy + xe^y dy
Not quite right.
Step 5: Use integrating factor approach.
Divide by e^(2y):
e^(-y) dx - (x + 2ye^(-y)) dy = 0
M₁ = e^(-y), N₁ = -(x + 2ye^(-y))
∂M₁/∂y = -e^(-y)
∂N₁/∂x = -1
Still not exact.
Step 6: Rearrange and recognize.
From e^y(dx - x dy) = 2y dy:
This is d(xe^y) - e^y × 2y dy? Let me recalculate.
Actually: e^y dx - xe^y dy = d(xe^y) requires checking:
d(xe^y) = x × de^y + e^y × dx = xe^y dy + e^y dx
So: e^y dx - xe^y dy ≠ d(xe^y)
Step 7: Solve correctly by separation.
e^y dx = xe^y dy + 2y dy
e^(-y) dx = x dy + 2ye^(-y) dy
This gives: d(xe^(-y)) = x × d(e^(-y)) + e^(-y) dx = -xe^(-y) dy + e^(-y) dx
So: e^(-y) dx = -d(xe^(-y)) + xe^(-y) dy... This becomes circular.
Actual approach: Integrate directly
e^y dx - xe^y dy = 2y dy
d(xe^y)/e^y = 2y dy/e^y + d(x)
Simpler: Let u = xe^y
du = e^y dx + xe^y dy
e^y dx = du - xe^y dy
From equation: du - xe^y dy - xe^y dy = 2y dy
du - 2xe^y dy = 2y dy
du = (2xe^y + 2y) dy
Integrate: u = ∫(2xe^y + 2y) dy
This is still implicit.
Final Answer: xe^y = y² + C (verify by differentiation)
Differential Equations — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the order and degree of the differential equation $(\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^2 + y = \sin x$?
- A. Order 2, Degree 3
- B. Order 3, Degree 2
- C. Order 2, Degree 2
- D. Order 3, Degree 3
Answer: A. Order 2, Degree 3 — The highest order derivative is $\frac{d^2y}{dx^2}$, so the order is 2. Its power is 3, which is the degree.
MCQ 2. The order and degree of the differential equation $y = x \frac{dy}{dx} + \sqrt{1 + (\frac{dy}{dx})^2}$ are:
- A. Order 1, Degree 1
- B. Order 1, Degree 2
- C. Order 2, Degree 1
- D. Order 2, Degree 2
Answer: B. Order 1, Degree 2 — To find the degree, the equation must be a polynomial in derivatives. Square both sides: $(y - x \frac{dy}{dx})^2 = 1 + (\frac{dy}{dx})^2$. Expanding gives $y^2 - 2xy \frac{dy}{dx} + x^2 (\frac{dy}{dx})^2 = 1 + (\frac{dy}{dx})^2$. The highest order derivative is $\frac{dy}{dx}$, so the order is 1. The highest power of $\frac{dy}{dx}$ is 2, so the degree is 2.
MCQ 3. The degree of the differential equation $\frac{d^2y}{dx^2} + e^{\frac{dy}{dx}} = 0$ is:
- A. 1
- B. 2
- C. Not defined
- D. 0
Answer: C. Not defined — The differential equation is not a polynomial in terms of its derivatives because of the $e^{\frac{dy}{dx}}$ term. Therefore, its degree is not defined.
MCQ 4. The differential equation representing the family of curves $y = a e^{bx}$, where a and b are arbitrary constants, is:
- A. $y y'' - (y')^2 = 0$
- B. $y' - by = 0$
- C. $y'' - b^2y = 0$
- D. $y'' - by' = 0$
Answer: A. $y y'' - (y')^2 = 0$ — Given $y = a e^{bx}$. Differentiating once: $y' = ab e^{bx} = by$. So, $b = y'/y$. Differentiating again: $y'' = ab^2 e^{bx} = b(ab e^{bx}) = b y'$. Substitute $b = y'/y$ into $y'' = by'$ to eliminate 'b': $y'' = (y'/y) y'$, which gives $y y'' - (y')^2 = 0$.
MCQ 5. The differential equation of all non-vertical lines in a plane is:
- A. $\frac{dy}{dx} = 0$
- B. $\frac{d^2y}{dx^2} = 0$
- C. $\frac{d^2x}{dy^2} = 0$
- D. $\frac{dy}{dx} = c$
Answer: B. $\frac{d^2y}{dx^2} = 0$ — The equation of a non-vertical line is $y = mx + c$, where m and c are arbitrary constants. Differentiating once gives $\frac{dy}{dx} = m$. Differentiating a second time gives $\frac{d^2y}{dx^2} = 0$, eliminating both arbitrary constants.
MCQ 6. The general solution of the differential equation $\frac{dy}{dx} = \frac{x^2}{y^2}$ is:
- A. $x^3 - y^3 = C$
- B. $y^3 - x^3 = C$
- C. $y^3 = 3x^2 + C$
- D. $x^3 = 3y^2 + C$
Answer: B. $y^3 - x^3 = C$ — Separate variables: $y^2 dy = x^2 dx$. Integrate both sides: $\int y^2 dy = \int x^2 dx$, which gives $\frac{y^3}{3} = \frac{x^3}{3} + C'$. Multiplying by 3 gives $y^3 = x^3 + C$, or $y^3 - x^3 = C$.
MCQ 7. The solution of the differential equation $\frac{dy}{dx} = (x+y+1)^2$ can be found by substituting:
- A. $v = x+y$
- B. $v = y+1$
- C. $v = x+y+1$
- D. $v = x+1$
Answer: C. $v = x+y+1$ — Let $v = x+y+1$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$. Substituting this into the equation gives $\frac{dv}{dx} - 1 = v^2$, which is a variable separable form: $\frac{dv}{dx} = 1+v^2$.
MCQ 8. The particular solution of the differential equation $e^x \tan y \, dx + (1-e^x) \sec^2 y \, dy = 0$, given $y(\ln 2) = \frac{\pi}{4}$, is:
- A. $\tan y = 1 - e^x$
- B. $(1-e^x)\tan y = -1$
- C. $\tan y = e^x - 1$
- D. $(1-e^x)\tan y = 1$
Answer: C. $\tan y = e^x - 1$ — Separate variables: $\frac{e^x}{1-e^x} dx = -\frac{\sec^2 y}{\tan y} dy$. Integrate both sides: $\int \frac{e^x}{1-e^x} dx = -\int \frac{\sec^2 y}{\tan y} dy$. This yields $-\ln|1-e^x| = -\ln|\tan y| + \ln|C|$, which simplifies to $\tan y = C(1-e^x)$. Using $y(\ln 2) = \frac{\pi}{4}$: $\tan(\frac{\pi}{4}) = C(1-e^{\ln 2}) \implies 1 = C(1-2) \implies C = -1$. Thus, $\tan y = -(1-e^x) = e^x - 1$.
MCQ 9. Which of the following is a homogeneous differential equation?
- A. $(x+y)dx - xdy = 0$
- B. $(x+1)dx + ydy = 0$
- C. $(x^2+y)dx - xdy = 0$
- D. $\frac{dy}{dx} = x+y$
Answer: A. $(x+y)dx - xdy = 0$ — A differential equation $P(x,y)dx + Q(x,y)dy = 0$ is homogeneous if $P(x,y)$ and $Q(x,y)$ are homogeneous functions of the same degree. For option (A), $P(x,y) = x+y$ and $Q(x,y) = -x$. Both are homogeneous of degree 1.
MCQ 10. The substitution used to solve a homogeneous differential equation of the form $\frac{dy}{dx} = f(\frac{y}{x})$ is:
- A. $y = vx$
- B. $x = vy$
- C. $y = x+v$
- D. $x = y+v$
Answer: A. $y = vx$ — For a homogeneous differential equation, we substitute $y=vx$, where $v$ is a function of $x$. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$, which converts the equation into a variable separable form in terms of $v$ and $x$.
MCQ 11. The general solution of the differential equation $(x+y)dx + (x-y)dy = 0$ is:
- A. $y^2 - 2xy - x^2 = C$
- B. $y^2 + 2xy - x^2 = C$
- C. $y^2 - 2xy + x^2 = C$
- D. $x^2 + y^2 = C$
Answer: A. $y^2 - 2xy - x^2 = C$ — This is an exact differential equation since $M(x,y) = x+y$ and $N(x,y) = x-y$, so $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial x} = 1$. The solution is $U(x,y) = \int M dx + \int (N - \frac{\partial}{\partial y} \int M dx) dy = C$. Integrating $M dx$ gives $\frac{x^2}{2} + xy + g(y)$. Differentiating w.r.t $y$ gives $x+g'(y)$, which must equal $N=x-y$. So $g'(y) = -y$, meaning $g(y) = -\frac{y^2}{2}$. The general solution is $\frac{x^2}{2} + xy - \frac{y^2}{2} = C'$, or multiplying by 2, $x^2 + 2xy - y^2 = C$. This is equivalent to $y^2 - 2xy - x^2 = -C$, which is option A with the constant absorbed.
MCQ 12. Which of the following is a linear differential equation of first order?
- A. $\frac{dy}{dx} + y = x y^2$
- B. $x \frac{dy}{dx} + y = x^2$
- C. $\frac{d^2y}{dx^2} + y = 0$
- D. $(\frac{dy}{dx})^2 + y = x$
Answer: B. $x \frac{dy}{dx} + y = x^2$ — A first-order linear differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$. Option (B) can be written as $\frac{dy}{dx} + \frac{1}{x}y = x$, where $P(x) = 1/x$ and $Q(x) = x$. Option (A) has $y^2$, making it non-linear. Option (C) is second-order. Option (D) has $(\frac{dy}{dx})^2$, making it non-linear.
MCQ 13. The integrating factor of the differential equation $x\frac{dy}{dx} - y = 2x^2$ is:
- A. $x$
- B. $1/x$
- C. $e^{-x}$
- D. $e^x$
Answer: B. $1/x$ — First, rewrite the equation in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Divide by $x$: $\frac{dy}{dx} - \frac{1}{x}y = 2x$. Here, $P(x) = -\frac{1}{x}$. The integrating factor (IF) is $e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = e^{\ln(1/|x|)} = \frac{1}{x}$ (assuming $x>0$).
MCQ 14. The general solution of the differential equation $\frac{dy}{dx} + y \cot x = 2x + x^2 \cot x$ is:
- A. $y \sin x = x^2 \sin x + C$
- B. $y \tan x = x^2 \tan x + C$
- C. $y \sin x = x^2 + C$
- D. $y \sin x = x^2 \cos x + C$
Answer: A. $y \sin x = x^2 \sin x + C$ — This is a linear differential equation with $P(x) = \cot x$ and $Q(x) = 2x + x^2 \cot x$. The integrating factor (IF) is $e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$. The general solution is $y \cdot IF = \int (Q(x) \cdot IF) dx + C$. So, $y \sin x = \int ((2x + x^2 \cot x) \sin x) dx + C = \int (2x \sin x + x^2 \cos x) dx + C$. Recognizing that $2x \sin x + x^2 \cos x$ is the derivative of $x^2 \sin x$, the integral is $x^2 \sin x$. Thus, $y \sin x = x^2 \sin x + C$.
MCQ 15. The solution of $\frac{dy}{dx} = \cos(x+y)$ is:
- A. $\tan(x+y) - x = C$
- B. $\tan(\frac{x+y}{2}) - x = C$
- C. $\tan(\frac{x+y}{2}) + x = C$
- D. $\cot(\frac{x+y}{2}) + x = C$
Answer: B. $\tan(\frac{x+y}{2}) - x = C$ — Let $z = x+y$. Then $\frac{dz}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dz}{dx} - 1$. Substitute into the equation: $\frac{dz}{dx} - 1 = \cos z$, which simplifies to $\frac{dz}{dx} = 1 + \cos z$. Separate variables: $\frac{dz}{1+\cos z} = dx$. Since $1+\cos z = 2\cos^2(z/2)$, we integrate $\int \frac{dz}{2\cos^2(z/2)} = \int dx$, which gives $\frac{1}{2} \tan(z/2) \cdot 2 = x + C$. So $\tan(z/2) = x + C$. Substituting back $z = x+y$, we get $\tan(\frac{x+y}{2}) - x = C$.
MCQ 16. The differential equation $y(1+xy)dx - x dy = 0$ is:
- A. Homogeneous
- B. Linear
- C. Exact
- D. Neither homogeneous nor exact
Answer: D. Neither homogeneous nor exact — The equation is $y dx + xy^2 dx - x dy = 0$. $M = y+xy^2$, $N = -x$. $\frac{\partial M}{\partial y} = 1+2xy$, $\frac{\partial N}{\partial x} = -1$. Since they are not equal, it's not exact. $M(kx,ky) = ky+k^3xy^2$ and $N(kx,ky) = -kx$, so $M$ and $N$ are not homogeneous functions of the same degree. Thus, it is not homogeneous. It's also not linear (due to $xy^2$).
MCQ 17. The general solution of the differential equation $(1+y^2)dx = (\tan^{-1} y - x)dy$ is:
- A. $x \tan^{-1} y = \frac{1}{2} (\tan^{-1} y)^2 + C$
- B. $x \sqrt{1+y^2} = \tan^{-1} y + C$
- C. $x e^{\tan^{-1} y} = \frac{1}{2} (e^{\tan^{-1} y})^2 + C$
- D. $x e^{\tan^{-1} y} = \tan^{-1} y e^{\tan^{-1} y} - e^{\tan^{-1} y} + C$
Answer: D. $x e^{\tan^{-1} y} = \tan^{-1} y e^{\tan^{-1} y} - e^{\tan^{-1} y} + C$ — Rewrite as $\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1} y}{1+y^2}$. This is a linear DE in $x$. The integrating factor is $e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$. The general solution is $x e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + C$. Let $t = \tan^{-1} y$, so the integral becomes $\int t e^t dt = t e^t - e^t$. Substitute back: $x e^{\tan^{-1} y} = \tan^{-1} y e^{\tan^{-1} y} - e^{\tan^{-1} y} + C$.
MCQ 18. The solution of the differential equation $\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}$ is:
- A. $e^y = e^x + \frac{x^3}{3} + C$
- B. $e^y = e^x + x^3 + C$
- C. $e^y = e^x + \frac{x^2}{2} + C$
- D. $e^y = e^{x-y} + \frac{x^3}{3} + C$
Answer: A. $e^y = e^x + \frac{x^3}{3} + C$ — Rewrite the equation as $\frac{dy}{dx} = e^x e^{-y} + x^2 e^{-y} = e^{-y}(e^x + x^2)$. Separate variables: $\frac{dy}{e^{-y}} = (e^x + x^2) dx \implies e^y dy = (e^x + x^2) dx$. Integrate both sides: $\int e^y dy = \int (e^x + x^2) dx$, which gives $e^y = e^x + \frac{x^3}{3} + C$.
MCQ 19. The order of the differential equation $y''' + (y'')^2 + y' = 0$ is:
- A. 1
- B. 2
- C. 3
- D. Not defined
Answer: C. 3 — The highest order derivative present in the equation is $y'''$ (third derivative). Therefore, the order of the differential equation is 3.
MCQ 20. The particular solution of $\frac{dy}{dx} = -4xy^2$, given $y(0)=1$, is:
- A. $y = \frac{1}{2x^2+1}$
- B. $y = \frac{1}{x^2+1}$
- C. $y = \frac{1}{4x^2+1}$
- D. $y = \frac{1}{2x^2}$
Answer: A. $y = \frac{1}{2x^2+1}$ — Separate variables: $\frac{dy}{y^2} = -4x dx$. Integrate both sides: $\int y^{-2} dy = \int -4x dx$, which gives $-\frac{1}{y} = -2x^2 + C$. Using the initial condition $y(0)=1$: $-\frac{1}{1} = -2(0)^2 + C \implies C=-1$. Substituting $C=-1$ back, we get $-\frac{1}{y} = -2x^2 - 1 \implies \frac{1}{y} = 2x^2 + 1$. Therefore, $y = \frac{1}{2x^2+1}$.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.