Solve (1 + x²) dy + 2xy dx = cot x dx. – Class 12 Mathematics | ConceptFirst

Question 33 of 50

Q. Solve (1 + x^2) dy + 2xy dx = cot x dx.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2019Subjective· 5mImportance★★★★★

66% · 33/50 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Rewrite as a first-order linear DE in $y$ , find the integrating factor $1+x^2$ , then integrate.

Given $(1+x^2)\,dy+2xy\,dx=\cot x\,dx$ . Divide by $dx$ and by $(1+x^2)$ :

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

This is linear in $y$ with $P(x)=\dfrac{2x}{1+x^2}$ , $Q(x)=\dfrac{\cot x}{1+x^2}$ .

Integrating factor:

$\text{IF}=e^{\int P\,dx}=e^{\int \frac{2x}{1+x^2}dx}=e^{\ln(1+x^2)}=1+x^2$

…

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL4 marks
$Q. Solve: e⁻ʸ sec²y dy = dx + x dy. OR Solve: x²(x dx + y dy) + 2y(x dy - y dx) = 0.$
›Reveal solutionSolution

Treat $x$ as the dependent variable (function of $y$ ) to get a linear first-order equation.

Given $e^{-y}\sec^2y\,dy = dx+x\,dy$ . Rearranging:

$e^{-y}\sec^2y\,dy - x\,dy = dx \;\Rightarrow\; \frac{dx}{dy} = e^{-y}\sec^2y - x \;\Rightarrow\; \frac{dx}{dy}+x = e^{-y}\sec^2y$

This is a linear differential equation in $x$ (as a function of $y$ ), of the form $\dfrac{dx}{dy}+Px=Q$ with $P=1,\ Q=e^{-y}\sec^2y$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL4 marks
$Q. Solve (x + 2y³) dy/dx = y, given x = 1, when y = -1. OR Solve the differential equation x dx + y dy + (x dy - y dx)/(x²+y²) = 0, given y = 1, when x = 1.$
›Reveal solutionSolution

Flip the equation to treat $x$ as a function of $y$ ; it becomes linear in $x$ , solvable with an integrating factor.

Given $(x+2y^3)\dfrac{dy}{dx} = y$ , given $x=1$ when $y=-1$ .

Rewrite as $\dfrac{dx}{dy} = \dfrac{x+2y^3}{y} = \dfrac{x}{y}+2y^2$ , i.e.:

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

This is a linear first-order ODE in $x$ (as a function of $y$ ), of the form $\dfrac{dx}{dy}+Px=Q$ with $P=-1/y$ , $Q=2y^2$ .

Integrating factor: $\displaystyle \text{I.F.} = e^{\int -1/y\,dy} = e^{-\ln y} = \frac1y$ .

Multiplying through and integrating: …

Start your 14-day free trial to see the full solution →

←Previous questionQ2 — Solve: (ex + 1) y dy - (y² + 1) ex dx = 0, given x = 0, y = 0. **OR** …→Next questionQ4 — Solve: dy/dx = e^(x-y) + x² e^(-y).