Board Exam-Pattern Questions

How This Bank Works

Note: These are original practice questions written in the CBSE/State Board pattern — matched to the style, difficulty and mark-distribution of recent board papers, not copied from any specific year's paper. Attempt each question on paper before reading the solution.

Differential Equations is one of the most predictable chapters on the board paper. Year after year, the questions cluster around five askable skills: reading off order and degree, verifying a given solution, solving by separation of variables, solving a homogeneous equation with $y = vx$, and solving a linear equation with an integrating factor. Almost every question is one of these five in light disguise.

How marks are typically distributed:

• 2-mark questions test definitions and one-step solves: order and degree, verification, writing down an integrating factor, or a short separable equation.
• 3-mark questions ask for a complete general or particular solution of a separable or linear equation — separation, integration, and the constant, all shown.
• 5-mark questions bring the heavier machinery: homogeneous equations with full substitution work, linear equations with by-parts integrals, or worded curve problems where you must build the equation before solving it.

Presentation rules that earn (and save) marks:

1. Name the method. Writing "this is a linear differential equation with $P = \cot x$, $Q = 4x\,\mathrm{cosec}\,x$" before computing anything signals to the examiner that you know exactly where you are going.
2. Show the separation or substitution line. The single line $\dfrac{dy}{h(y)} = g(x)\,dx$ or $y = vx \Rightarrow \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ typically carries its own mark.
3. Never drop the constant. Add $C$ at the moment of integration, not at the end. In particular-solution questions, the substitution of the initial condition is a dedicated step with a dedicated mark.
4. Present the final answer in the asked form. If the question says "show that $\dots = C$", massage your answer to that exact shape — logs merged, constants renamed.

2-Mark Questions

Q1. Write the order and the degree (if defined) of the differential equation $\left(\dfrac{d^2y}{dx^2}\right)^2 + \cos\left(\dfrac{dy}{dx}\right) = 0$.

Solution:

1. Order: the highest derivative present is $\dfrac{d^2y}{dx^2}$, so the order is $2$.
2. Degree: the equation contains $\cos\left(\dfrac{dy}{dx}\right)$ — a derivative sitting inside a cosine — so the equation is not a polynomial in its derivatives. The degree is not defined.

Answer: order $2$; degree not defined.

Q2. Find the general solution of $\dfrac{dy}{dx} = \sqrt{4 - y^2}$, where $-2 < y < 2$.

Solution:

1. Separate: $\dfrac{dy}{\sqrt{4 - y^2}} = dx$.
2. Integrate: $\displaystyle\int \dfrac{dy}{\sqrt{2^2 - y^2}} = \int dx \Rightarrow \sin^{-1}\dfrac{y}{2} = x + C$.
3. Solve for $y$: $y = 2\sin(x + C)$.

Answer: $y = 2\sin(x + C)$.

Q3. Write the integrating factor of the differential equation $x\dfrac{dy}{dx} - y = x^2$.

Solution:

1. Standard form first (divide by $x$): $\dfrac{dy}{dx} - \dfrac{y}{x} = x$, so $P = -\dfrac{1}{x}$.
2. I.F.: $e^{\int P\,dx} = e^{-\log|x|} = \dfrac{1}{x}$.

Answer: I.F. $= \dfrac{1}{x}$. (For completeness: multiplying through gives $\dfrac{d}{dx}\left(\dfrac{y}{x}\right) = 1$, so $y = x^2 + Cx$.)

Q4. Verify that $y = e^{-3x}$ is a solution of the differential equation $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0$.

Solution:

1. Differentiate twice: $\dfrac{dy}{dx} = -3e^{-3x}$, $\dfrac{d^2y}{dx^2} = 9e^{-3x}$.
2. Substitute: $9e^{-3x} + (-3e^{-3x}) - 6e^{-3x} = (9 - 3 - 6)e^{-3x} = 0$. ✓

Answer: the left side vanishes identically, so $y = e^{-3x}$ is a solution.

3-Mark Questions

Q5. Find the general solution of the differential equation $\dfrac{dy}{dx} + \sqrt{\dfrac{1 - y^2}{1 - x^2}} = 0$.

Solution:

1. Separate: $\dfrac{dy}{dx} = -\dfrac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \Rightarrow \dfrac{dy}{\sqrt{1 - y^2}} = -\dfrac{dx}{\sqrt{1 - x^2}}$.
2. Integrate both sides: $\sin^{-1} y = -\sin^{-1} x + C$.
3. Present: $\sin^{-1} y + \sin^{-1} x = C$.

Answer: $\sin^{-1} x + \sin^{-1} y = C$.

Q6. Find the equation of the curve passing through the point $\left(0, \dfrac{\pi}{4}\right)$ whose differential equation is $\sin x \cos y\,dx + \cos x \sin y\,dy = 0$.

Solution:

1. Separate (divide by $\cos x \cos y$): $\tan x\,dx + \tan y\,dy = 0$.
2. Integrate: $-\log|\cos x| - \log|\cos y| = \log C_1$, i.e. $\log|\cos x \cos y| = -\log C_1$, so $\cos x \cos y = C$.
3. Apply the point: at $x = 0$, $y = \dfrac{\pi}{4}$: $\cos 0 \cdot \cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$, so $C = \dfrac{1}{\sqrt{2}}$.

Answer: $\cos x \cos y = \dfrac{1}{\sqrt{2}}$.

Q7. Find the particular solution of the differential equation $\left(1 + e^{2x}\right)dy + \left(1 + y^2\right)e^x\,dx = 0$, given that $y = 1$ when $x = 0$.

Solution:

1. Separate: $\dfrac{dy}{1 + y^2} = -\dfrac{e^x\,dx}{1 + e^{2x}}$.
2. Integrate — on the right put $t = e^x$, $dt = e^x dx$: $\tan^{-1} y = -\tan^{-1}(e^x) + C$.
3. Apply the condition: $\tan^{-1} 1 + \tan^{-1}(e^0) = C \Rightarrow \dfrac{\pi}{4} + \dfrac{\pi}{4} = C = \dfrac{\pi}{2}$.

Answer: $\tan^{-1} y + \tan^{-1}\left(e^x\right) = \dfrac{\pi}{2}$.

Q8. Find the particular solution of the differential equation $\dfrac{dy}{dx} + y\cot x = 4x\,\mathrm{cosec}\,x$ $(x \neq 0)$, given that $y = 0$ when $x = \dfrac{\pi}{2}$.

Solution:

1. Classify: linear, with $P = \cot x$, $Q = 4x\,\mathrm{cosec}\,x$.
2. I.F.: $e^{\int \cot x\,dx} = e^{\log|\sin x|} = \sin x$.
3. Solve: $y\sin x = \displaystyle\int 4x\,\mathrm{cosec}\,x \cdot \sin x\,dx = \int 4x\,dx = 2x^2 + C$.
4. Apply the condition: $0 = 2\cdot\dfrac{\pi^2}{4} + C \Rightarrow C = -\dfrac{\pi^2}{2}$.

Answer: $y\sin x = 2x^2 - \dfrac{\pi^2}{2}$.

5-Mark Questions

Q9. Solve the differential equation $y e^{\frac{x}{y}}\,dx = \left(x e^{\frac{x}{y}} + y^2\right)dy$ $(y \neq 0)$.

Solution:

1. Reorient: $\dfrac{dx}{dy} = \dfrac{x e^{x/y} + y^2}{y e^{x/y}}$. The ratio $\dfrac{x}{y}$ dominates — try $x = vy$, so $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$.
2. Substitute: $v + y\dfrac{dv}{dy} = \dfrac{vy\,e^v + y^2}{y\,e^v} = v + \dfrac{y}{e^v}$.
3. Cancel and separate: $y\dfrac{dv}{dy} = \dfrac{y}{e^v} \Rightarrow e^v\,dv = dy$.
4. Integrate: $e^v = y + C$.
5. Return to $x, y$: $e^{\frac{x}{y}} = y + C$.

Answer: $e^{\frac{x}{y}} = y + C$.

Q10. Find a particular solution of the differential equation $(x - y)(dx + dy) = dx - dy$, given that $y = -1$ when $x = 0$.

Solution:

1. Substitute $x - y = t$: then $dt = dx - dy$ and $dx + dy = 2dx - dt$. The equation reads $t(dx + dy) = dt$.
2. Rearrange in $t$: from $t\,(dx + dy) = dt$ and $dx + dy = \dfrac{dt}{t}$, note also $dx + dy = 2dx - dt$, so $2dx = dt + \dfrac{dt}{t} = \left(1 + \dfrac{1}{t}\right)dt$… a cleaner route: write the original as $\dfrac{dx - dy}{dx + dy} = t$, i.e. $\dfrac{dt}{d(x+y)} = x - y$.
3. Separate: with $s = x + y$: $\dfrac{dt}{t} = ds$.
4. Integrate: $\log|t| = s + C$, i.e. $\log|x - y| = x + y + C$.
5. Apply the condition $(x, y) = (0, -1)$: $\log|0 - (-1)| = 0 + (-1) + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$.

Answer: $\log|x - y| = x + y + 1$.

Q11. Solve the differential equation $\left(\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}} - \dfrac{y}{\sqrt{x}}\right)\dfrac{dx}{dy} = 1$ $(x \neq 0)$.

Solution:

1. Flip to standard form: $\dfrac{dy}{dx} = \dfrac{e^{-2\sqrt{x}}}{\sqrt{x}} - \dfrac{y}{\sqrt{x}}$, i.e. $\dfrac{dy}{dx} + \dfrac{1}{\sqrt{x}}\,y = \dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}$ — linear with $P = \dfrac{1}{\sqrt{x}}$.
2. I.F.: $\displaystyle\int \dfrac{dx}{\sqrt{x}} = 2\sqrt{x}$, so I.F. $= e^{2\sqrt{x}}$.
3. Solve: $y\,e^{2\sqrt{x}} = \displaystyle\int e^{2\sqrt{x}}\cdot\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}\,dx = \int \dfrac{dx}{\sqrt{x}} = 2\sqrt{x} + C$.

Answer: $y\,e^{2\sqrt{x}} = 2\sqrt{x} + C$. The exponentials annihilating each other inside the integral is the intended reward for a correct I.F.

Q12. Find a particular solution of the differential equation $(x + 1)\dfrac{dy}{dx} = 2e^{-y} - 1$, given that $y = 0$ when $x = 0$.

Solution:

1. Separate: $\dfrac{dy}{2e^{-y} - 1} = \dfrac{dx}{x + 1}$.
2. Clean the left side (multiply top and bottom by $e^y$): $\dfrac{e^y\,dy}{2 - e^y} = \dfrac{dx}{x + 1}$.
3. Integrate — left side with $u = 2 - e^y$, $du = -e^y dy$: $-\log\left|2 - e^y\right| = \log|x + 1| + C_1$, so $(2 - e^y)(x + 1) = C$.
4. Apply the condition $(0, 0)$: $(2 - 1)(1) = C = 1$.
5. Solve for $y$: $2 - e^y = \dfrac{1}{x + 1} \Rightarrow e^y = \dfrac{2x + 1}{x + 1}$.

Answer: $y = \log\left|\dfrac{2x + 1}{x + 1}\right|$, $x \neq -1$.