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 14 on this concept.

1. GUJCET 2025Set 031 markMCQ
   Q.The Integrating Factor of the differential equation $x \cdot \frac{dy}{dx} + 2y = x^2,\ (x \neq 0)$ is _____ (A) $\frac{1}{x^2}$ (B) $e^{-x}$ (C) $e^{-y}$ (D) $x^2$

   ›Reveal solutionSolution

   Put in the form $\tfrac{dy}{dx}+P(x)y=Q(x)$ and use $\text{IF}=e^{\int P\,dx}$.

   Dividing by $x$: $\tfrac{dy}{dx}+\tfrac{2}{x}y=x$, so $P=\tfrac{2}{x}$ and …

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2. GUJCET 2020Set 071 markMCQ
   Q.Integrating factor of differential equation $(\tan^{-1} y - x)\,dy = (1 + y^2)\,dx$ is ________. (A) $e^{1 + y^2}$ (B) $e^y$ (C) $e^{\tan^{-1} x}$ (D) $e^{\tan^{-1} y}$

   ›Reveal solutionSolution

   Treat $x$ as the dependent variable; the integrating factor is $e^{\int\frac{dy}{1+y^2}}=e^{\tan^{-1}y}$.

   Concept: $(\tan^{-1}y-x)\,dy=(1+y^2)\,dx$ gives

   $\frac{dx}{dy}=\frac{\tan^{-1}y-x}{1+y^2}\ \Rightarrow\ \frac{dx}{dy}+\frac{1}{1+y^2}\,x=\frac{\tan^{-1}y}{1+y^2}.$ …

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3. GUJCET 2024Set 131 markMCQ
   Q.The Integrating Factor of the differential equation $(\tan^{-1} y - x)\,dy = (1 + y^2)\,dx$ is __________. (A) $e^{\tan^{-1} y}$ (B) $\frac{1}{1+y^2}$ (C) $e^{\frac{1}{1+y^2}}$ (D) $\tan^{-1} y$

   ›Reveal solutionSolution

   Treated as linear in $x$, the integrating factor is $e^{\int \frac{dy}{1+y^2}} = e^{\tan^{-1}y}$.

   Steps. From $(\tan^{-1}y - x)\,dy = (1+y^2)\,dx$,

   $\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1+y^2} \ \Rightarrow\ \frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2}.$ …

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4. GUJCET 2023Set 091 markMCQ
   Q.The integrating factor of the differential equation $\frac{dy}{dx} + y\tan x = \sec x$ is : (A) $\tan x$ (B) $e^{\sec x}$ (C) $\cos x$ (D) $\sec x$

   ›Reveal solutionSolution

   [!TLDR] Max $R$ = longest length ÷ smallest area, i.e. current along the 10 cm length entering the $1\times\tfrac12$ cm faces.

   Concept

   Resistance of a uniform conductor is $R = \dfrac{\rho L}{A}$, where $L$ is the length along the current direction and $A$ is the perpendicular cross-sectional area. To maximise $R$, choose the connection giving the greatest $L$ and the smallest $A$.

   Solution

   The rod is $10\,\text{cm}\times1\,\text{cm}\times\tfrac12\,\text{cm}$. The three ways to connect across opposite faces:

   • Across $10\times1$ cm faces: $L = \tfrac12$ cm, $A = 10$ cm$^2$ → $R \propto 0.5/10 = 0.05$. …

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5. GUJCET 2022Set 081 markMCQ
   Q.The integrating factor of the differential equation $x\frac{dy}{dx}-y=x^2$ is ______. (A) $e^{-x}$ (B) $\frac{1}{x}$ (C) $e^{x}$ (D) $x$

   ›Reveal solutionSolution

   Divide by $x$ to get $\dfrac{dy}{dx}-\dfrac{y}{x}=x$; then $\text{IF}=e^{\int(-1/x)dx}=\tfrac1x$.

   Concept. $x\dfrac{dy}{dx}-y=x^2\Rightarrow\dfrac{dy}{dx}-\dfrac{1}{x}y=x$. Here $P(x)=-\tfrac1x$, so …

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6. GUJCET 2019Set 171 markMCQ
   Q.The integrating factor (I.F.) of differential equation $\dfrac{dy}{dx}(1+x) - xy = 1-x$ is $\underline{\hspace{2cm}}$. (A) $(x-1)e^{-x}$ (B) $(1+x)e^{-x}$ (C) $(1+x)e^{x}$ (D) $(1-x)e^{-x}$

   ›Reveal solutionSolution

   Reduce to $\dfrac{dy}{dx}-\dfrac{x}{1+x}y=\dfrac{1-x}{1+x}$; I.F. $=e^{\int P\,dx}$.

   Concept. Divide by $(1+x)$: $P(x) = -\dfrac{x}{1+x}$.

   Steps.

   • $\displaystyle\int \frac{x}{1+x}dx = \int\left(1-\frac{1}{1+x}\right)dx = x-\log(1+x).$ …

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7. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.The general solution of the differential equation $e^x\,dy + (ye^x+2x)\,dx = 0$ is ____.

   (a) $y\cdot e^x + x^2 = C$

   (b) $y\cdot e^x - x^2 = C$

   (c) $x\cdot e^x + y^2 = C$

   (d) $x\cdot e^x - y^2 = C$

   ›Reveal solutionSolution

   Divide through by $e^x$ to get a linear ODE in $y$, then solve with an integrating factor of $1$ (equation is already exact after simplification).

   $e^x dy+(ye^x+2x)dx=0\Rightarrow \dfrac{dy}{dx}+y=-2xe^{-x}$. Integrating factor $=e^{\int1\,dx}=e^x$.

   …

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8. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.The integrating factor of the differential equation $(\tan^{-1}y - x)\,dy = (1+y^2)\,dx$ is ____.

   (a) $e^{\tan^{-1}x}$

   (b) $e^{\tan^{-1}y}$

   (c) $e^{-\tan^{-1}x}$

   (d) $e^{-\tan^{-1}y}$

   ›Reveal solutionSolution

   Rewrite as a linear equation in $x$ (treating $y$ as the independent variable) and find its integrating factor.

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

   …

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9. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.The integrating factor of the differential equation $(1 - y^2)\dfrac{dx}{dy} + yx = ay$, $(-1 < y < 1)$, = ____.

   (a) $\dfrac{1}{y^2 - 1}$

   (b) $\dfrac{1}{1 - y^2}$

   (c) $\dfrac{1}{\sqrt{y^2 - 1}}$

   (d) $\dfrac{1}{\sqrt{1 - y^2}}$

   ›Reveal solutionSolution

   Write the equation in standard linear form $\dfrac{dx}{dy}+Px=Q$ and compute $\text{I.F.}=e^{\int P\,dy}$.

   Divide by $(1-y^2)$: $\dfrac{dx}{dy}+\dfrac{y}{1-y^2}x=\dfrac{ay}{1-y^2}$, so $P=\dfrac{y}{1-y^2}$.

   Let $u=1-y^2$, $du=-2y\,dy$: $\displaystyle\int\dfrac{y}{1-y^2}dy=-\dfrac12\ln|1-y^2|$.

   …

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10. GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
    Q.The integrating factor of the differential equation $x\frac{dy}{dx} + 2y = x^2$ $(x \ne 0)$ is ______.

    (a) $2\log x$

    (b) $\log x$

    (c) $\frac{2}{x}$

    (d) $x^2$

    ›Reveal solutionSolution

    Write the equation in standard linear form $\frac{dy}{dx}+Py=Q$ and use $IF=e^{\int P\,dx}$.

    Divide by $x$: $\frac{dy}{dx}+\frac{2}{x}y=x$, so $P=\frac{2}{x}$.

    …

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11. GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
    Q.The integrating factor of the differential equation $x\frac{dy}{dx} + 2y = x^2 \log x$ is ___.

    (a) $e^{2x}$

    (b) $x^2$

    (c) $e^x$

    (d) $x$

    ›Reveal solutionSolution

    Put in standard linear form and compute IF $= e^{\int P\,dx}$.

    $x\dfrac{dy}{dx}+2y = x^2\log x \Rightarrow \dfrac{dy}{dx} + \dfrac2x\,y = x\log x.$

    …

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12. GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
    Q.Integrating factor of the differential equation $y\,dx-(x+2y^2)\,dy=0$ is ___.

    (a) $-1/y$

    (b) $-y$

    (c) $y$

    (d) $1/y$

    ›Reveal solutionSolution

    Rearrange into the linear form $\dfrac{dx}{dy}+P(y)x=Q(y)$ (treating $x$ as the dependent variable), then compute $I.F.=e^{\int P\,dy}$.

    $y\,dx-(x+2y^2)dy=0 \Rightarrow y\,dx=(x+2y^2)dy \Rightarrow \dfrac{dx}{dy}=\dfrac{x}{y}+2y \Rightarrow \dfrac{dx}{dy}-\dfrac1y\,x=2y$. …

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