Questions and Answers (Board Exam)

Introduction to Board Exam PYQs

This section features 29 highly important Previous Year Questions (PYQs) from board-style examinations, aligned with trends up to 2026. In board exams, method and presentation are crucial. Always clearly identify the type of differential equation first — whether it is about order and degree, formation of differential equation, variable separable, homogeneous, or linear differential equation.

For Linear Differential Equations, explicitly state the standard form, identify $P$ and $Q$, and write the Integrating Factor (I.F.) separately before solving. For equations solved by integration, do not forget to introduce the arbitrary constant $C$ immediately after integrating.

Question 1 [CBSE 2026]

Find the sum of the order and degree of the differential equation: $\frac{d}{dx}\left( \left(\frac{dy}{dx}\right)^3 \right) = 0$.

Solution: Step 1: Differentiate the given expression using the chain rule. $$\frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^3\right] = 3\left(\frac{dy}{dx}\right)^2 \frac{d^2y}{dx^2}$$ So the equation becomes: $$3\left(\frac{dy}{dx}\right)^2 \frac{d^2y}{dx^2} = 0$$

Step 2: Identify the highest order derivative. The highest derivative appearing is $\frac{d^2y}{dx^2}$. Therefore, the order = 2.

Step 3: Find the degree. The equation is polynomial in derivatives, and the highest order derivative $\frac{d^2y}{dx^2}$ occurs to the first power. Therefore, the degree = 1.

Step 4: Find the sum. $$\text{Order} + \text{Degree} = 2 + 1 = 3$$

Answer: 3

Question 2 [CBSE 2025]

Find the order and degree of the differential equation: $\left(\frac{d^2s}{dt^2}\right)^2 + 3\left(\frac{ds}{dt}\right)^3 + 4 = 0$.

Solution: Step 1: Find the highest order derivative. The derivatives present are $\frac{ds}{dt}$ and $\frac{d^2s}{dt^2}$. The highest order derivative is $\frac{d^2s}{dt^2}$. Hence, the order = 2.

Step 2: Check whether the equation is polynomial in derivatives. Yes. All derivatives appear with non-negative integer powers, and none is inside a transcendental function.

Step 3: Determine the degree. The highest order derivative $\frac{d^2s}{dt^2}$ is raised to the power 2. Hence, the degree = 2.

Answer: Order = 2, Degree = 2

Question 3 [CBSE 2024]

Write the differential equation representing the family of curves $y = A e^{2x} + B e^{-2x}$, where $A$ and $B$ are arbitrary constants.

Solution: Step 1: Since there are two arbitrary constants $A$ and $B$, we must differentiate twice to eliminate them.

Given: $$y = A e^{2x} + B e^{-2x}$$

Step 2: Differentiate once. $$\frac{dy}{dx} = 2A e^{2x} - 2B e^{-2x}$$

Step 3: Differentiate again. $$\frac{d^2y}{dx^2} = 4A e^{2x} + 4B e^{-2x}$$ But from the original equation, $$A e^{2x} + B e^{-2x} = y$$ So, $$\frac{d^2y}{dx^2} = 4y$$

Step 4: Rearrange in standard differential equation form. $$\frac{d^2y}{dx^2} - 4y = 0$$

Answer: $\frac{d^2y}{dx^2} - 4y = 0$

Question 4 [CBSE 2023]

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

Solution: Step 1: This is a variable separable differential equation because all $y$ terms can be brought to one side and all $x$ terms to the other side.

Separate the variables: $$\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$$

Step 2: Integrate both sides. $$\int \frac{1}{1+y^2} \, dy = \int \frac{1}{1+x^2} \, dx$$

Step 3: Use the standard integral formula: $$\int \frac{1}{1+z^2}dz = \tan^{-1} z$$ So we get: $$\tan^{-1} y = \tan^{-1} x + C$$

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

Question 5 [CBSE 2026]

Find the general solution of: $\frac{dy}{dx} = e^{x+y}$.

Solution: Step 1: Rewrite the right-hand side as a product of a function of $x$ and a function of $y$: $$\frac{dy}{dx} = e^x e^y$$ This is a variable separable differential equation.

Step 2: Separate the variables. $$e^{-y}dy = e^x dx$$

Step 3: Integrate both sides. $$\int e^{-y}dy = \int e^x dx$$

Step 4: Evaluate the integrals. $$-e^{-y} = e^x + C$$ Multiply through by $-1$: $$e^{-y} = -e^x + C_1$$ Equivalently, writing the constant in a cleaner combined form, $$e^x + e^{-y} = K$$ where $K$ is an arbitrary constant.

Answer: $e^x + e^{-y} = K$

Question 6 [CBSE 2025]

Solve: $(e^x + e^{-x}) dy - (e^x - e^{-x}) dx = 0$.

Solution: Step 1: Rearrange the equation. $$(e^x + e^{-x})dy = (e^x - e^{-x})dx$$ So, $$dy = \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$$

Step 2: Integrate both sides. $$\int dy = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$$

Step 3: Observe that the numerator is the derivative of the denominator: $$\frac{d}{dx}(e^x + e^{-x}) = e^x - e^{-x}$$ Therefore, $$y = \ln|e^x + e^{-x}| + C$$ Since $e^x + e^{-x} > 0$ for all real $x$, absolute value is not necessary, though it is mathematically acceptable.

Answer: $y = \log(e^x + e^{-x}) + C$

Question 7 [CBSE 2024]

Find the general solution of: $\log\left(\frac{dy}{dx}\right) = 3x + 4y$.

Solution: Step 1: Convert the logarithmic form into exponential form. $$\frac{dy}{dx} = e^{3x+4y} = e^{3x} e^{4y}$$ This is now variable separable.

Step 2: Separate the variables. $$e^{-4y} dy = e^{3x} dx$$

Step 3: Integrate both sides. $$\int e^{-4y} dy = \int e^{3x} dx$$

Step 4: Evaluate the integrals. $$-\frac{1}{4}e^{-4y} = \frac{1}{3}e^{3x} + C$$ Multiply throughout by 12: $$-3e^{-4y} = 4e^{3x} + C_1$$ Rearranging, $$4e^{3x} + 3e^{-4y} = K$$ where $K$ is an arbitrary constant.

Answer: $4e^{3x} + 3e^{-4y} = K$

Question 8 [CBSE 2023]

Solve the differential equation: $\sqrt{1-x^2} dy + \sqrt{1-y^2} dx = 0$.

Solution: Step 1: Separate variables. $$\sqrt{1-x^2}dy = -\sqrt{1-y^2}dx$$ $$\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$$

Step 2: Integrate both sides. $$\int \frac{dy}{\sqrt{1-y^2}} = -\int \frac{dx}{\sqrt{1-x^2}}$$

Step 3: Use the standard formula: $$\int \frac{1}{\sqrt{1-z^2}}dz = \sin^{-1} z$$ Thus, $$\sin^{-1} y = -\sin^{-1} x + C$$ Rearranging, $$\sin^{-1} x + \sin^{-1} y = C$$

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

Question 9 [CBSE 2026]

Solve: $x \sqrt{1+y^2} dx + y \sqrt{1+x^2} dy = 0$.

Solution: Step 1: Rearrange to separate variables. $$y\sqrt{1+x^2}dy = -x\sqrt{1+y^2}dx$$ $$\frac{y}{\sqrt{1+y^2}}dy = -\frac{x}{\sqrt{1+x^2}}dx$$

Step 2: Integrate both sides. $$\int \frac{y}{\sqrt{1+y^2}}dy = -\int \frac{x}{\sqrt{1+x^2}}dx$$

Step 3: Use substitution on both sides. For the left side, let $u = 1+y^2$, then $du = 2y dy$. So, $$\int \frac{y}{\sqrt{1+y^2}}dy = \sqrt{1+y^2}$$ Similarly, $$\int \frac{x}{\sqrt{1+x^2}}dx = \sqrt{1+x^2}$$

Step 4: Therefore, $$\sqrt{1+y^2} = -\sqrt{1+x^2} + C$$ or, $$\sqrt{1+x^2} + \sqrt{1+y^2} = C$$

Answer: $\sqrt{1+x^2} + \sqrt{1+y^2} = C$

Question 10 [CBSE 2022]

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

Solution: Step 1: Separate the variables. $$\frac{dy}{y^2} = -4x dx$$

Step 2: Integrate both sides. $$\int y^{-2}dy = \int -4x dx$$ $$-\frac{1}{y} = -2x^2 + C$$ Multiply by $-1$: $$\frac{1}{y} = 2x^2 + C_1$$

Step 3: Apply the initial condition $y(0)=1$. $$1 = 0 + C_1 \implies C_1 = 1$$

Step 4: Substitute the value of the constant. $$\frac{1}{y} = 2x^2 + 1$$ Hence, $$y = \frac{1}{2x^2+1}$$

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

Question 11 [CBSE 2025]

Solve: $\frac{dy}{dx} = 1 - x + y - xy$.

Solution: Step 1: Factor the right-hand side. $$1 - x + y - xy = (1-x) + y(1-x) = (1-x)(1+y)$$ So the differential equation becomes: $$\frac{dy}{dx} = (1-x)(1+y)$$

Step 2: Separate variables. $$\frac{dy}{1+y} = (1-x)dx$$

Step 3: Integrate both sides. $$\int \frac{dy}{1+y} = \int (1-x)dx$$

Step 4: Evaluate the integrals. $$\log|1+y| = x - \frac{x^2}{2} + C$$

Answer: $\log|1+y| = x - \frac{x^2}{2} + C$

Question 12 [CBSE 2024]

Solve: $y \log y dx - x dy = 0$.

Solution: Step 1: Rearrange the equation. $$y\log y \, dx = x \, dy$$ $$\frac{dx}{x} = \frac{dy}{y\log y}$$

Step 2: Integrate both sides. $$\int \frac{dx}{x} = \int \frac{dy}{y\log y}$$

Step 3: For the right side, use substitution: Let $$t = \log y \implies dt = \frac{1}{y}dy$$ Then, $$\int \frac{dy}{y\log y} = \int \frac{dt}{t} = \log|t| + C = \log|\log y| + C$$

Step 4: So, $$\log|x| = \log|\log y| + C$$ Write $C = \log|K|$ for simplification: $$\log|x| = \log|K\log y|$$ Hence, $$x = K\log y$$

Answer: $x = C \log y$

Question 13 [CBSE 2026]

Solve the homogeneous differential equation: $(x^2+xy)dy = (x^2+y^2)dx$.

Solution: Step 1: Rewrite the equation in differential form. $$\frac{dy}{dx} = \frac{x^2+y^2}{x^2+xy}$$ Since numerator and denominator are homogeneous expressions of degree 2, the equation is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = \frac{1+v^2}{1+v}$$

Step 3: Simplify. $$x\frac{dv}{dx} = \frac{1+v^2}{1+v} - v = \frac{1-v}{1+v}$$ Thus, $$\frac{1+v}{1-v}dv = \frac{dx}{x}$$

Step 4: Simplify the integrand before integrating. $$\frac{1+v}{1-v} = -1 + \frac{2}{1-v}$$ So, $$\int \left(-1 + \frac{2}{1-v}\right) dv = \int \frac{dx}{x}$$

Step 5: Integrate. $$-v - 2\log|1-v| = \log|x| + C$$

Step 6: Substitute back $v = y/x$. $$-\frac{y}{x} - 2\log\left|1-\frac{y}{x}\right| = \log|x| + C$$ Now, $$1-\frac{y}{x} = \frac{x-y}{x}$$ Hence, $$\log\left|1-\frac{y}{x}\right| = \log|x-y| - \log|x|$$ Substituting, $$-\frac{y}{x} - 2\log|x-y| + 2\log|x| = \log|x| + C$$ Therefore, $$-\frac{y}{x} - 2\log|x-y| + \log|x| = C$$

Answer: $-\frac{y}{x} - 2\log|x-y| + \log|x| = C$

Question 14 [CBSE 2023]

Solve: $x dy - y dx = \sqrt{x^2+y^2} dx$.

Solution: Step 1: Rearrange to isolate $\frac{dy}{dx}$. $$xdy = (y + \sqrt{x^2+y^2})dx$$ $$\frac{dy}{dx} = \frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2}$$ This is a homogeneous differential equation.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = v + \sqrt{1+v^2}$$

Step 3: Simplify and separate. $$x\frac{dv}{dx} = \sqrt{1+v^2}$$ $$\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$$

Step 4: Integrate. $$\log|v+\sqrt{1+v^2}| = \log|x| + \log|C|$$ $$v+\sqrt{1+v^2} = Cx$$

Step 5: Substitute back $v = y/x$. $$\frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2} = Cx$$ Multiply through by $x$: $$y + \sqrt{x^2+y^2} = Cx^2$$

Answer: $y + \sqrt{x^2+y^2} = Cx^2$

Question 15 [CBSE 2025]

Solve: $x \frac{dy}{dx} = y - x \tan\left(\frac{y}{x}\right)$.

Solution: Step 1: Divide by $x$: $$\frac{dy}{dx} = \frac{y}{x} - \tan\left(\frac{y}{x}\right)$$ This is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = v - \tan v$$

Step 3: Simplify. $$x\frac{dv}{dx} = -\tan v$$ $$\cot v \, dv = -\frac{dx}{x}$$

Step 4: Integrate. $$\int \cot v \, dv = -\int \frac{dx}{x}$$ $$\log|\sin v| = -\log|x| + \log|C|$$ So, $$\log|\sin v| = \log\left|\frac{C}{x}\right|$$ $$\sin v = \frac{C}{x}$$

Step 5: Replace $v = y/x$. $$x\sin\left(\frac{y}{x}\right) = C$$

Answer: $x \sin\left(\frac{y}{x}\right) = C$

Question 16 [CBSE 2026]

Solve: $(x^2-y^2)dx + 2xy dy = 0$.

Solution: Step 1: Rearrange in differential equation form. $$2xy \, dy = (y^2-x^2)dx$$ $$\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$$ This is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = \frac{v^2-1}{2v}$$

Step 3: Simplify. $$x\frac{dv}{dx} = \frac{v^2-1}{2v} - v = -\frac{1+v^2}{2v}$$ So, $$\frac{2v}{1+v^2}dv = -\frac{dx}{x}$$

Step 4: Integrate. $$\int \frac{2v}{1+v^2}dv = -\int \frac{dx}{x}$$ $$\log(1+v^2) = -\log|x| + \log|C|$$ Hence, $$1+v^2 = \frac{C}{x}$$

Step 5: Replace $v = y/x$. $$1 + \frac{y^2}{x^2} = \frac{C}{x}$$ Multiply by $x^2$: $$x^2 + y^2 = Cx$$

Answer: $x^2+y^2 = Cx$

Question 17 [CBSE 2024]

Solve: $y dx + x \log\left(\frac{y}{x}\right) dy - 2x dy = 0$.

Solution: Step 1: Rearrange to express $\frac{dx}{dy}$. $$y \, dx = x\left(2 - \log\frac{y}{x}\right)dy$$ $$\frac{dx}{dy} = \frac{x}{y}\left(2 - \log\frac{y}{x}\right)$$ This is homogeneous in the form $\frac{dx}{dy} = f(x/y)$.

Step 2: Put $$x = vy \implies \frac{dx}{dy} = v + y\frac{dv}{dy}$$ Also, $$\frac{y}{x} = \frac{1}{v}, \qquad \log\left(\frac{y}{x}\right) = -\log v$$ So the equation becomes: $$v + y\frac{dv}{dy} = v(2 + \log v)$$

Step 3: Simplify. $$y\frac{dv}{dy} = v(1 + \log v)$$ Now separate the variables: $$\frac{dv}{v(1+\log v)} = \frac{dy}{y}$$

Step 4: Integrate. Let $$u = 1 + \log v \implies du = \frac{1}{v}dv$$ Then, $$\int \frac{du}{u} = \int \frac{dy}{y}$$ $$\log|1+\log v| = \log|y| + C$$

Step 5: Remove logarithms. $$1 + \log v = Cy$$ Substitute back $v = x/y$: $$1 + \log\left(\frac{x}{y}\right) = Cy$$

Answer: $1 + \log\left(\frac{x}{y}\right) = Cy$

Question 18 [CBSE 2022]

Solve: $(x^3+y^3)dy - x^2y dx = 0$.

Solution: Step 1: Rearrange the equation. $$\frac{dy}{dx} = \frac{x^2y}{x^3+y^3}$$ This is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = \frac{v}{1+v^3}$$

Step 3: Simplify. $$x\frac{dv}{dx} = \frac{v}{1+v^3} - v = -\frac{v^4}{1+v^3}$$ So, $$\frac{1+v^3}{v^4}dv = -\frac{dx}{x}$$ $$\left(v^{-4} + v^{-1}\right)dv = -\frac{dx}{x}$$

Step 4: Integrate. $$\int v^{-4}dv + \int v^{-1}dv = -\int \frac{dx}{x}$$ $$-\frac{1}{3v^3} + \log|v| = -\log|x| + C$$ Now, $$\log|v| + \log|x| = \log|vx| = \log|y|$$ Hence, $$\log|y| - \frac{1}{3v^3} = C$$ Since $$v = \frac{y}{x} \implies \frac{1}{v^3} = \frac{x^3}{y^3}$$ we get $$\log|y| - \frac{x^3}{3y^3} = C$$

Answer: $\log|y| - \frac{x^3}{3y^3} = C$

Question 19 [CBSE 2025]

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

Solution: Step 1: Rearrange to isolate $\frac{dy}{dx}$. $$2x^2\frac{dy}{dx} = 2xy + y^2$$ $$\frac{dy}{dx} = \frac{2xy+y^2}{2x^2}$$ This is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = \frac{2v+v^2}{2} = v + \frac{v^2}{2}$$

Step 3: Simplify and separate. $$x\frac{dv}{dx} = \frac{v^2}{2}$$ $$\frac{dv}{v^2} = \frac{dx}{2x}$$

Step 4: Integrate. $$\int v^{-2}dv = \frac{1}{2}\int \frac{dx}{x}$$ $$-\frac{1}{v} = \frac{1}{2}\log|x| + C$$

Step 5: Replace $v = y/x$. $$-\frac{x}{y} = \frac{1}{2}\log|x| + C$$

Step 6: Apply the condition $y=2$ when $x=1$. $$-\frac{1}{2} = 0 + C \implies C = -\frac{1}{2}$$

Step 7: Substitute back. $$-\frac{x}{y} = \frac{1}{2}\log|x| - \frac{1}{2}$$ Multiply by $-2$: $$\frac{2x}{y} = 1 - \log|x|$$ Hence, $$y = \frac{2x}{1 - \log|x|}$$

Answer: $y = \frac{2x}{1 - \log|x|}$

Question 20 [CBSE 2026]

Solve: $\frac{dy}{dx} + 2y = e^{3x}$.

Solution: Step 1: This is a linear differential equation in the form: $$\frac{dy}{dx} + Py = Q$$ Here, $$P = 2, \qquad Q = e^{3x}$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int 2dx} = e^{2x}$$

Step 3: Apply the linear differential equation formula. $$y e^{2x} = \int e^{3x}e^{2x}dx + C$$ $$y e^{2x} = \int e^{5x}dx + C$$

Step 4: Integrate. $$y e^{2x} = \frac{e^{5x}}{5} + C$$ So, $$y = \frac{e^{3x}}{5} + Ce^{-2x}$$

Answer: $y = \frac{e^{3x}}{5} + Ce^{-2x}$

Question 21 [CBSE 2024]

Solve: $\frac{dy}{dx} + y \cot x = 2x + x^2 \cot x$.

Solution: Step 1: This is a linear differential equation with $$P = \cot x, \qquad Q = 2x + x^2\cot x$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \cot x \, dx} = e^{\log(\sin x)} = \sin x$$

Step 3: Apply the formula. $$y\sin x = \int (2x + x^2\cot x)\sin x \, dx + C$$ $$y\sin x = \int (2x\sin x + x^2\cos x)dx + C$$

Step 4: Notice that the integrand is the derivative of $x^2\sin x$: $$\frac{d}{dx}(x^2\sin x) = 2x\sin x + x^2\cos x$$ Therefore, $$y\sin x = x^2\sin x + C$$

Answer: $y \sin x = x^2 \sin x + C$

Question 22 [CBSE 2023]

Solve: $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$.

Solution: Step 1: Divide the equation by $x\log x$. $$\frac{dy}{dx} + \frac{1}{x\log x}y = \frac{2}{x^2}$$ Thus, $$P = \frac{1}{x\log x}, \qquad Q = \frac{2}{x^2}$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \frac{1}{x\log x}dx}$$ Let $$t = \log x \implies dt = \frac{1}{x}dx$$ Then, $$\int \frac{1}{x\log x}dx = \int \frac{dt}{t} = \log(\log x)$$ Therefore, $$\text{I.F.} = e^{\log(\log x)} = \log x$$

Step 3: Apply the formula. $$y\log x = \int \frac{2}{x^2}\log x \, dx + C$$ But this is not simplified correctly if we stop there. Since the formula requires $Q \cdot \text{I.F.}$, we compute: $$Q \cdot \text{I.F.} = \frac{2}{x^2} \cdot \log x$$ So, $$y\log x = \int \frac{2\log x}{x^2}dx + C$$

Step 4: Integrate by parts. Let $$u = \log x, \quad dv = \frac{2}{x^2}dx$$ Then, $$du = \frac{1}{x}dx, \quad v = -\frac{2}{x}$$ Thus, $$\int \frac{2\log x}{x^2}dx = -\frac{2\log x}{x} + \int \frac{2}{x^2}dx$$ $$= -\frac{2\log x}{x} - \frac{2}{x}$$

Therefore, $$y\log x = -\frac{2\log x}{x} - \frac{2}{x} + C$$ or, $$y\log x = -\frac{2}{x}(\log x + 1) + C$$

Answer: $y \log x = -\frac{2}{x}(\log x + 1) + C$

Question 23 [CBSE 2025]

Solve: $(1+x^2)dy + 2xy dx = \cot x dx$ ($x \neq 0$).

Solution: Step 1: Divide by $dx$ and then by $(1+x^2)$: $$\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{\cot x}{1+x^2}$$ This is linear with $$P = \frac{2x}{1+x^2}, \qquad Q = \frac{\cot x}{1+x^2}$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \frac{2x}{1+x^2}dx} = e^{\log(1+x^2)} = 1+x^2$$

Step 3: Apply the formula. $$y(1+x^2) = \int \frac{\cot x}{1+x^2}(1+x^2)dx + C$$ $$y(1+x^2) = \int \cot x \, dx + C$$

Step 4: Integrate. $$\int \cot x \, dx = \log|\sin x|$$ Therefore, $$y(1+x^2) = \log|\sin x| + C$$

Answer: $y(1+x^2) = \log|\sin x| + C$

Question 24 [CBSE 2022]

Solve: $\frac{dy}{dx} + y \sec x = \tan x$.

Solution: Step 1: This is linear with $$P = \sec x, \qquad Q = \tan x$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \sec x dx} = e^{\log(\sec x + \tan x)} = \sec x + \tan x$$

Step 3: Apply the formula. $$y(\sec x + \tan x) = \int \tan x(\sec x + \tan x)dx + C$$ $$= \int (\sec x\tan x + \tan^2x)dx + C$$ Use the identity: $$\tan^2x = \sec^2x - 1$$ So, $$y(\sec x + \tan x) = \int (\sec x\tan x + \sec^2x - 1)dx + C$$

Step 4: Integrate termwise. $$y(\sec x + \tan x) = \sec x + \tan x - x + C$$

Answer: $y(\sec x + \tan x) = \sec x + \tan x - x + C$

Question 25 [CBSE 2026]

Solve: $(x+y) \frac{dy}{dx} = 1$.

Solution: Step 1: It is not linear in $y$, but if we take reciprocal, it becomes linear in $x$. $$\frac{dx}{dy} = x + y$$ Thus, $$\frac{dx}{dy} - x = y$$ This is linear in $x$ with $$P_1 = -1, \qquad Q_1 = y$$

Step 2: Find the Integrating Factor with respect to $y$. $$\text{I.F.} = e^{\int -1 \, dy} = e^{-y}$$

Step 3: Apply the formula. $$xe^{-y} = \int ye^{-y}dy + C$$

Step 4: Integrate by parts. Let $$u = y, \quad dv = e^{-y}dy$$ Then, $$du = dy, \quad v = -e^{-y}$$ Hence, $$\int ye^{-y}dy = -ye^{-y} - e^{-y}$$ So, $$xe^{-y} = -ye^{-y} - e^{-y} + C$$ Multiply by $e^y$: $$x = -y - 1 + Ce^y$$

Answer: $x + y + 1 = Ce^y$

Question 26 [CBSE 2024]

Solve: $(1+y^2)dx = (\tan^{-1}y - x)dy$.

Solution: Step 1: Rearrange into linear form in $x$. $$\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1+y^2}$$ $$\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1}y}{1+y^2}$$ This is linear with $$P_1 = \frac{1}{1+y^2}, \qquad Q_1 = \frac{\tan^{-1}y}{1+y^2}$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \frac{1}{1+y^2}dy} = e^{\tan^{-1}y}$$

Step 3: Apply the formula. $$xe^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2} e^{\tan^{-1}y} dy + C$$ Let $$t = \tan^{-1}y \implies dt = \frac{1}{1+y^2}dy$$ Then, $$\int t e^t dt = te^t - e^t$$ Thus, $$xe^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C$$ So, $$x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$$

Answer: $x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$

Question 27 [CBSE 2025]

Solve the IVP: $\frac{dy}{dx} + \frac{y}{x} = e^x$, $x>0$, given $y(1)=1$.

Solution: Step 1: This is linear with $$P = \frac{1}{x}, \qquad Q = e^x$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \frac{1}{x}dx} = e^{\log x} = x$$

Step 3: Apply the formula. $$yx = \int xe^x dx + C$$

Step 4: Integrate by parts. We know: $$\int xe^x dx = xe^x - e^x$$ So, $$yx = e^x(x-1) + C$$

Step 5: Apply the initial condition $y(1)=1$. $$1\cdot 1 = e(1-1) + C \implies C = 1$$

Step 6: Therefore, $$yx = e^x(x-1) + 1$$

Answer: $yx = e^x(x-1) + 1$

Question 28 [CBSE 2023]

Solve: $\cos^2x \frac{dy}{dx} + y = \tan x$.

Solution: Step 1: Divide by $\cos^2x$. $$\frac{dy}{dx} + y\sec^2x = \tan x\sec^2x$$ This is linear with $$P = \sec^2x, \qquad Q = \tan x\sec^2x$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \sec^2x dx} = e^{\tan x}$$

Step 3: Apply the formula. $$ye^{\tan x} = \int \tan x\sec^2x e^{\tan x} dx + C$$ Let $$t = \tan x \implies dt = \sec^2x dx$$ Then, $$\int te^t dt = e^t(t-1)$$ So, $$ye^{\tan x} = e^{\tan x}(\tan x - 1) + C$$

Step 4: Divide by $e^{\tan x}$. $$y = \tan x - 1 + Ce^{-\tan x}$$

Answer: $y = \tan x - 1 + Ce^{-\tan x}$