Q.Solve the following differential equation: $(1 + x^2) dy + 2xy \ dx = \cot x \ dx \quad (x \ne 0)$

Showing the 12 most recent of 15 on this concept.

1. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.The integrating factor of the differential equation $(3\sin x\cos x)\,dy=(1+3y\sin^2 x)\,dx$, where $0<x<\dfrac{\pi}{2}$, is (A) $\sec x$ (B) $\sin x$ (C) $\tan x$ (D) $\cos x$ (E) $\cot x$

   ›Reveal solutionSolution

   Put in linear form; the coefficient of $y$ gives IF $=\cos x$.

   From $(3\sin x\cos x)dy=(1+3y\sin^2 x)dx$:

   $\frac{dy}{dx}=\frac{1+3y\sin^2 x}{3\sin x\cos x}=\frac{1}{3\sin x\cos x}+\frac{3\sin^2 x}{3\sin x\cos x}y.$

   The $y$-coefficient is $\frac{\sin x}{\cos x}=\tan x$, so …

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2. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The integrating factor of the differential equation $(1+x^2)dy=(1-2xy)dx$ is (A) $x^2+1$ (B) $\log_e(x^2+1)$ (C) $\dfrac{x}{x^2+1}$ (D) $x(x^2+1)$ (E) $\log_e|x|$

   ›Reveal solutionSolution

   Put in standard linear form $\dfrac{dy}{dx}+Py=Q$ and compute $e^{\int P\,dx}$.

   Rewrite $(1+x^{2})dy=(1-2xy)dx$ as

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

   Here $P=\dfrac{2x}{1+x^{2}}$, so …

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3. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.The integrating factor of the differential equation $\sin x\, dy = \dfrac{1}{2}(\sin 2x + 2y\cos x)\, dx$ is (A) $\sec x$ (B) $\sin x$ (C) $\tan x$ (D) $\cos x$ (E) $\csc x$

   ›Reveal solutionSolution

   The equation reduces to y' - (cot x) y = cos x, giving integrating factor e^{-integral cot x dx} = csc x.

   Concept and Intuition

   Rewrite the equation in standard linear form dy/dx + P(x)y = Q(x); the integrating factor is e^{integral P dx}.

   Step-by-Step Solution

   1. sin x dy = (1/2)(sin2x + 2y cos x) dx; using sin2x = 2 sin x cos x, the RHS = (sin x cos x + y cos x) dx.
   2. So sin x (dy/dx) = sin x cos x + y cos x, i.e. dy/dx = cos x + y cot x. …

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4. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.The general solution of the differential equation $2y\tan x+\dfrac{dy}{dx}=5\sin x$ is (A) $y=5\sec x+C\sec^2 x$ (B) $y=5+C\cos x$ (C) $y=5\cos x+C$ (D) $y=5\cos x+C\cos^2 x$ (E) $y=5\sec^2 x+C\sec x$

   ›Reveal solutionSolution

   Solve the linear ODE with integrating factor $\sec^2 x$.

   Rewrite as $\frac{dy}{dx}+2\tan x\,y=5\sin x$. Integrating factor:

   $\mu=e^{\int 2\tan x\,dx}=e^{2\log\sec x}=\sec^2 x.$

   Then

   $\frac{d}{dx}\left(y\sec^2 x\right)=5\sin x\sec^2 x=5\frac{\sin x}{\cos^2 x}.$ …

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5. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.The general solution of the differential equation $(x^2 y^2 + y)dx - (x - 2x^3 y)dy = 0$ is (A) $x^2 y^2 - \dfrac{y}{x} = C$ (B) $x^3 y + \dfrac{x}{y} = C$ (C) $xy^2 + \dfrac{y}{x} = C$ (D) $xy^2 - \dfrac{y}{x} = C$ (E) $x^2 y + \dfrac{y}{x} = C$

   ›Reveal solutionSolution

   The general solution is $xy^2-\frac{y}{x}=C$.

   Concept and Intuition

   Regroup terms into recognizable exact differentials $d(xy^2)$ and $d(y/x)$.

   Step-by-Step Solution

   1. Expand: $(x^2y^2+y)dx-(x-2x^3y)dy=0\Rightarrow x^2y^2\,dx+y\,dx-x\,dy+2x^3y\,dy=0$.
   2. Note $d(xy^2)=y^2\,dx+2xy\,dy$, so $x^2y^2\,dx+2x^3y\,dy=x^2\,d(xy^2)$.
   3. Note $d(y/x)=\frac{x\,dy-y\,dx}{x^2}$, so $y\,dx-x\,dy=-x^2\,d(y/x)$.
   4. Equation: $x^2\,d(xy^2)-x^2\,d(y/x)=0\Rightarrow d(xy^2)-d(y/x)=0$. …

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6. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The solution of the differential equation $(y + x^2)\, dx = x\, dy, x > 0$ is a curve which passes through the point $(1, 0)$. The equation of the curve is (A) $y = x(x + 1)$ (B) $y = x(x - 1)$ (C) $y = x^2(x - 1)$ (D) $y = x^2(x + 1)$ (E) $y = x(x^2 - 1)$

   ›Reveal solutionSolution

   Rewrite as a first-order linear ODE $y'-y/x=x$, use integrating factor $1/x$, then apply the point $(1,0)$.

   From $(y+x^2)dx=x\,dy$: $x\frac{dy}{dx}=y+x^2\Rightarrow \frac{dy}{dx}-\frac{y}{x}=x$.

   Integrating factor $=e^{-\int dx/x}=\frac1x$, so $\frac{d}{dx}\Big(\frac{y}{x}\Big)=1$. …

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7. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The general solution of the differential equation $x^3 \dfrac{dy}{dx} + 3x^2 y = \cos x$ is (A) $y = \dfrac{\sin x}{x^3} + C$ (B) $y = \dfrac{\sin x}{x^3} + Cx$ (C) $y = \dfrac{\sin x}{x^2} + C$ (D) $y = \dfrac{\sin x + C}{x^2}$ (E) $y = \dfrac{\sin x + C}{x^3}$

   ›Reveal solutionSolution

   Recognize the exact derivative $\dfrac{d}{dx}(x^3y)$.

   Since $\dfrac{d}{dx}(x^3y)=x^3\dfrac{dy}{dx}+3x^2y$, the equation $x^3\dfrac{dy}{dx}+3x^2y=\cos x$ becomes

   $\frac{d}{dx}(x^3y)=\cos x.$ …

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8. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.The integrating factor of the differential equation $xy' + 2y - 7x^3 = 0$ is (A) $\log|x|$ (B) $x^2$ (C) $\dfrac{1}{x^2}$ (D) $\dfrac{1}{2}\log|x|$ (E) $x$

   ›Reveal solutionSolution

   The integrating factor is $x^2$.

   Concept and Intuition

   A linear ODE $y'+P(x)y=Q(x)$ has integrating factor $e^{\int P\,dx}$. First put the equation in standard form by dividing through by the coefficient of $y'$.

   Step-by-Step Solution

   1. $xy'+2y-7x^3=0\Rightarrow y'+\dfrac{2}{x}y=7x^2$.
   2. Here $P(x)=\dfrac{2}{x}$, so $\int P\,dx=2\log|x|$. …

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9. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The integrating factor of the differential equation $2\, dy = (y + \cos x)\, dx$ is (A) $e^{-2x}$ (B) $e^{-x/2}$ (C) $e^{2x}$ (D) $e^{x/2}$ (E) $-e^{x/2}$

   ›Reveal solutionSolution

   Put in standard linear form and read off the integrating factor.

   From $2\,dy=(y+\cos x)\,dx$:

   $2\frac{dy}{dx}=y+\cos x\ \Rightarrow\ \frac{dy}{dx}-\frac12 y=\frac12\cos x.$ …

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10. KEAM 2024Set eng-2024-06064 marksMCQ
    Q.The integrating factor of the differential equation $x\frac{dy}{dx}+2y=xe^{x}$ is (A) $\log_{e} x$ (B) $\log_{e} 2x$ (C) $x$ (D) $x^{2}$ (E) $2x$

    ›Reveal solutionSolution

    Dividing by $x$ gives $P(x)=\frac{2}{x}$, so the integrating factor is $x^2$.

    Divide $x\frac{dy}{dx}+2y=xe^x$ by $x$: $\frac{dy}{dx}+\frac{2}{x}y=e^x$. …

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11. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.The general solution of $\dfrac{dy}{dx} + y = 5$ is (A) $-\log|5 - y| = x + C$ (B) $-\log|5 - y| = e^x + C$ (C) $(5 - y)^2 = 2x + C$ (D) $y = \log|x + C|$ (E) $\log|x| + C$

    ›Reveal solutionSolution

    The equation is separable: $\frac{dy}{5-y}=dx$, which integrates to $-\log|5-y|=x+C$.

    Write $\frac{dy}{dx}=5-y$, so $\frac{dy}{5-y}=dx$. …

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12. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.The integrating factor of the differential equation $4x\,dy - e^{-2y}dy + dx = 0$ is (A) $e^{-2y}$ (B) $e^{2x^2}$ (C) $e^{4y}$ (D) $e^{-4y}$ (E) $x^4$

    ›Reveal solutionSolution

    The integrating factor is $e^{4y}$.

    Concept and Intuition

    Rewrite the equation as linear in $x$ with respect to $y$; the integrating factor is $e^{\int P\,dy}$.

    Step-by-Step Solution

    1. $4x\,dy-e^{-2y}\,dy+dx=0\Rightarrow (4x-e^{-2y})\,dy+dx=0$.
    2. Divide by $dy$: $\frac{dx}{dy}+4x=e^{-2y}$.
    3. This is linear with $P(y)=4$.
    4. Integrating factor $=e^{\int4\,dy}=e^{4y}$.

    Common Mistakes …

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