Q.The Integrating Factor of the differential equation $(1 - y^2)\dfrac{dx}{dy} + yx = ay \ (-1 < y < 1)$ is (A) $\dfrac{1}{y^2 - 1}$ (B) $\dfrac{1}{\sqrt{y^2 - 1}}$ (C) $\dfrac{1}{1 - y^2}$ (D) $\dfrac{1}{\sqrt{1 - y^2}}$

Showing the 12 most recent of 33 on this concept.

1. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.An integrating factor of the differential equation $(x^2+1)\dfrac{dy}{dx} + xy = x^3$ is (A) $\dfrac{x}{1+x^2}$ (B) $\dfrac{1}{2}\log(1+x^2)$ (C) $\sqrt{1+x^2}$ (D) $e^{\log(1+x^2)}$

   ›Reveal solutionSolution

   Writing the ODE in standard linear form $\frac{dy}{dx}+Py=Q$ and computing $e^{\int P\,dx}$ gives the integrating factor $\sqrt{1+x^2}$.

   Concept and Intuition

   A first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$ is solved by multiplying through by the integrating factor $\mu(x)=e^{\int P(x)\,dx}$, chosen precisely so that $\dfrac{d}{dx}\big[\mu(x)y\big]=\mu(x)Q(x)$. So the entire method hinges on correctly identifying $P(x)$ after dividing the equation into standard form.

   Step-by-Step Solution

   1. Start with $(x^2+1)\dfrac{dy}{dx}+xy=x^3$.
   2. Divide by $(x^2+1)$ to reach standard form: $\dfrac{dy}{dx}+\dfrac{x}{x^2+1}y=\dfrac{x^3}{x^2+1}$.
   3. Here $P(x)=\dfrac{x}{x^2+1}$.
   4. Compute $\int P(x)\,dx=\displaystyle\int \dfrac{x}{x^2+1}dx=\dfrac12\log(x^2+1)$ (substituting $u=x^2+1$). …

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2. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Integrating factor of the differential equation $\sin x \dfrac{dy}{dx} - y\cos x = 1$ is (A) $\sin x$ (B) $\cos x$ (C) $\sec x$ (D) $\mathrm{cosec}\, x$

   ›Reveal solutionSolution

   Rewriting the equation in standard linear form and computing $e^{\int P\,dx}$ gives the integrating factor $\csc x$ — (D).

   Concept and Intuition

   A first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$ has integrating factor $e^{\int P(x)dx}$. The given equation must first be divided through so the coefficient of $\dfrac{dy}{dx}$ is exactly $1$.

   Step-by-Step Solution

   1. Given: $\sin x\dfrac{dy}{dx}-y\cos x=1$.
   2. Divide by $\sin x$: $\dfrac{dy}{dx}-\dfrac{\cos x}{\sin x}y=\dfrac1{\sin x}$, i.e. $\dfrac{dy}{dx}-(\cot x)y=\csc x$.
   3. Here $P(x)=-\cot x$.
   4. Integrating factor $=e^{\int -\cot x\,dx}=e^{-\log|\sin x|}=(\sin x)^{-1}=\csc x$.

   Common Mistakes …

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3. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\log y$ is an integrating factor of $\dfrac{dx}{dy}+P(y)x=Q(y)$, then $P(y)=$ (A) $\dfrac{1}{y+\log y}$ (B) $\dfrac{y}{\log y}$ (C) $\dfrac{\log y}{y}$ (D) $\dfrac{1}{y\log y}$

   ›Reveal solutionSolution

   Working backward from the known integrating factor $\log y=e^{\int P(y)dy}$ and differentiating recovers $P(y)=\dfrac{1}{y\log y}$.

   Concept and Intuition

   For a linear ODE $\dfrac{dx}{dy}+P(y)x=Q(y)$ (linear in $x$, with $y$ as the independent variable), the integrating factor is $\mathrm{IF}=e^{\int P(y)\,dy}$. If we're told what the integrating factor equals, we can reverse-engineer $P(y)$ by taking logs and then differentiating.

   Step-by-Step Solution

   1. Standard fact: $\mathrm{IF}=e^{\int P(y)\,dy}$.
   2. Given $\mathrm{IF}=\log y$ (i.e. $\log y$), set $e^{\int P(y)dy}=\log y$.
   3. Take the natural log of both sides: $\displaystyle\int P(y)\,dy=\log(\log y)$.
   4. Differentiate both sides with respect to $y$ (by the Fundamental Theorem of Calculus, the left side gives back $P(y)$): $P(y)=\dfrac{d}{dy}\left[\log(\log y)\right]$.
   5. By the chain rule: $\dfrac{d}{dy}\log(\log y)=\dfrac{1}{\log y}\cdot\dfrac{d}{dy}(\log y)=\dfrac{1}{\log y}\cdot\dfrac1y=\dfrac{1}{y\log y}$. …

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4. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If $x\,dy+(y+y^2x)\,dx=0$ and $y=1$ at $x=1$, then (A) $y=\dfrac{x}{1+\log x}$ (B) $y=\dfrac{1+\log x}{x}$ (C) $y=x(1+\log x)$ (D) $y=\dfrac{1}{x(1+\log x)}$

   ›Reveal solutionSolution

   A Bernoulli equation in disguise; the substitution $v=1/y$ linearizes it, and the initial condition $y(1)=1$ pins down the constant to give $y=\dfrac{1}{x(1+\log x)}$.

   Concept and Intuition

   An equation of the form $\dfrac{dy}{dx}+P(x)y=Q(x)y^n$ is a Bernoulli equation. Dividing by $y^n$ and substituting $v=y^{1-n}$ converts it into a linear first-order ODE in $v$, which can then be solved with the standard integrating-factor method.

   Step-by-Step Solution

   1. Given $x\,dy+(y+y^2x)\,dx=0$. Divide by $dx$: $x\dfrac{dy}{dx}+y+y^2x=0$.
   2. Divide by $x$: $\dfrac{dy}{dx}+\dfrac{y}{x}=-y^2$ — a Bernoulli equation with $n=2$.
   3. Divide throughout by $y^2$: $y^{-2}\dfrac{dy}{dx}+\dfrac{1}{xy}=-1$.
   4. Let $v=y^{-1}$, so $\dfrac{dv}{dx}=-y^{-2}\dfrac{dy}{dx}$, i.e. $y^{-2}\dfrac{dy}{dx}=-\dfrac{dv}{dx}$.
   5. Substitute: $-\dfrac{dv}{dx}+\dfrac{v}{x}=-1 \Rightarrow \dfrac{dv}{dx}-\dfrac{v}{x}=1$, a linear ODE in $v$.
   6. Integrating factor $=e^{\int -1/x\,dx}=e^{-\log x}=\dfrac{1}{x}$.
   7. $\dfrac{d}{dx}\!\left(\dfrac{v}{x}\right)=\dfrac{1}{x} \Rightarrow \dfrac{v}{x}=\log x+C \Rightarrow v=x(\log x+C)$. …

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5. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The differential equation $f'(y)\dfrac{dy}{dx} + P(x)f(y) = x^3$ is reduced to linear differential equation by substituting $Z = f(y)$. If the integrating factor of the reduced linear equation is $e^{x^2}$, then the solution of the given differential equation is (A) $x^2 + Ae^{-x^2} - 2f(y) = 1$ (B) $f(y) = \dfrac{1}{2}(x^2-1) + Ce^{x^2}$ (C) $x^2 + Ae^{x^2} + 2f(y) = 1$ (D) $f(y) = (x^3 + Ce^{x^2})$

   ›Reveal solutionSolution

   After the substitution $Z=f(y)$, this becomes a standard linear first-order DE with known integrating factor $e^{x^2}$; solving and rearranging gives $x^2+Ae^{-x^2}-2f(y)=1$.

   Concept and Intuition

   Bernoulli-style substitutions like $Z=f(y)$ are used precisely to convert a DE that's nonlinear/awkward in $y$ into a genuinely linear DE in the new variable $Z$, which can then be solved by the standard integrating-factor method.

   Step-by-Step Solution

   1. With $Z=f(y)$, $\dfrac{dZ}{dx}=f'(y)\dfrac{dy}{dx}$, so the given equation becomes $\dfrac{dZ}{dx}+P(x)Z=x^3$ — linear in $Z$.
   2. The integrating factor is $e^{\int P(x)dx}$, given as $e^{x^2}$, so $\int P\,dx=x^2$ (i.e. $P(x)=2x$, though we don't need this explicitly).
   3. Standard linear-DE solution: $Z\cdot e^{x^2}=\displaystyle\int x^3 e^{x^2}dx+C$.
   4. Compute $\displaystyle\int x^3e^{x^2}dx$: let $u=x^2$, $du=2x\,dx$, so $x^3e^{x^2}dx=\tfrac12\,u e^u\,du$. Using $\int ue^udu=e^u(u-1)$: result is $\tfrac12e^{x^2}(x^2-1)$.
   5. So $Z e^{x^2}=\tfrac12e^{x^2}(x^2-1)+C$, giving $Z=f(y)=\tfrac12(x^2-1)+Ce^{-x^2}$. …

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6. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.Solve the following Differential equation. $(x^2+1)\dfrac{dy}{dx} + 4xy = \dfrac{1}{x^2+1}$ (A) $y(x^2-1)^2 = x + c$ (B) $y(x^2+1)^2 = x + c$ (C) $y(x^2+1)^2 = x^2 + c$ (D) $y(x^2-1)^2 = x^2 + c$

   ›Reveal solutionSolution

   Divide through to get standard linear-ODE form, find the integrating factor, and the equation collapses to a total derivative.

   Concept and Intuition

   A first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$ is solved by multiplying through by the integrating factor $\mu=e^{\int P\,dx}$, which makes the left side exactly $\dfrac{d}{dx}(\mu y)$.

   Step-by-Step Solution

   1. Divide the given equation by $(x^2+1)$: $\dfrac{dy}{dx}+\dfrac{4x}{x^2+1}y=\dfrac1{(x^2+1)^2}$.
   2. $P(x)=\dfrac{4x}{x^2+1}$; integrating factor $\mu=e^{\int\frac{4x}{x^2+1}dx}=e^{2\log(x^2+1)}=(x^2+1)^2$.
   3. Multiply the standard-form equation by $\mu$: $(x^2+1)^2\dfrac{dy}{dx}+4x(x^2+1)y=1$.
   4. The left side is exactly $\dfrac{d}{dx}\left[y(x^2+1)^2\right]$. …

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7. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.The integrating factor of the linear differential equation $\dfrac{dy}{dx} + P(x)y = Q(x)$ is a solution of the differential equation (A) $\dfrac{dy}{dx} - P(x)y = 0$ (B) $\dfrac{dy}{dx} + P(x)y = 0$ (C) $\dfrac{dy}{dx} - \dfrac{y}{x} = P(x)$ (D) $\dfrac{dy}{dx} + \dfrac{x}{y} = P(x)$

   ›Reveal solutionSolution

   The integrating factor $\mu=e^{\int P\,dx}$ is itself a solution of the homogeneous equation $y'=P(x)y$.

   Concept and Intuition

   The whole point of the integrating factor is that $\dfrac{d}{dx}\big(\mu y\big)=\mu\dfrac{dy}{dx}+\mu P(x)y$ should equal $\mu\left(\dfrac{dy}{dx}+P(x)y\right)$, which requires $\dfrac{d\mu}{dx}=\mu P(x)$ — i.e. $\mu$ itself solves the first-order linear homogeneous ODE $y'-P(x)y=0$.

   Step-by-Step Solution

   1. The integrating factor is defined as $\mu(x)=e^{\int P(x)\,dx}$.
   2. Differentiate: $\mu'(x) = P(x)\,e^{\int P\,dx} = P(x)\mu(x)$.
   3. Rearranged: $\mu'(x) - P(x)\mu(x)=0$. …

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8. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If $y=y(x)$ is a particular solution of $\sqrt{1-x^2}\frac{dy}{dx}+\frac{2x}{\sqrt{1-x^2}}y=x$, $y(0)=1$, then $y\left(\frac{1}{2}\right)=$ (A) $\frac{\sqrt3}{2}$ (B) $\frac{1}{4}$ (C) $\frac{1}{2}$ (D) $0$

   ›Reveal solutionSolution

   This is a linear first-order ODE; its integrating factor is $\dfrac{1}{1-x^2}$, leading to the clean solution $y=\sqrt{1-x^2}$, so $y(1/2)=\sqrt3/2$.

   Concept and Intuition

   After dividing by $\sqrt{1-x^2}$, the equation becomes the standard linear form $y'+P(x)y=Q(x)$ with $P(x)=\dfrac{2x}{1-x^2}$. Since $P$ is (up to sign) the logarithmic derivative of $1-x^2$, the integrating factor collapses to a simple power of $(1-x^2)$, which is the usual sign that a linear ODE has been set up correctly.

   Step-by-Step Solution

   1. Divide the given equation by $\sqrt{1-x^2}$: $\dfrac{dy}{dx}+\dfrac{2x}{1-x^2}y=\dfrac{x}{\sqrt{1-x^2}}$.
   2. Integrating factor: $\mu(x)=\exp\!\left(\int\dfrac{2x}{1-x^2}dx\right)=\exp(-\log(1-x^2))=\dfrac{1}{1-x^2}$.
   3. Multiply through: $\dfrac{d}{dx}\!\left[\dfrac{y}{1-x^2}\right]=\dfrac{x}{(1-x^2)^{3/2}}$.
   4. Integrate the RHS (substitute $w=1-x^2$, $dw=-2x\,dx$): $\int\dfrac{x\,dx}{(1-x^2)^{3/2}}=\dfrac{1}{\sqrt{1-x^2}}+C$. …

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9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The general solution of the differential equation $(1+y^2) + (x - e^{\tan^{-1}y})\frac{dy}{dx} = 0$ is (A) $xe^{\tan^{-1}y} = \tan^{-1}y + c$ (B) $x^2e^{2\tan^{-1}y} = e^{\tan^{-1}y} + c$ (C) $(x-2) = ce^{-\tan^{-1}y}$ (D) $2xe^{\tan^{-1}y} = e^{2\tan^{-1}y} + c$

   ›Reveal solutionSolution

   This is a first-order linear ODE that becomes linear in $x$ (not $y$) once you recognise the $\tan^{-1}y$ integrating factor. The answer is (D).

   Concept and Intuition

   When the equation is not linear in $y$ but treating $x$ as the dependent variable (function of $y$) makes it linear, we should switch roles: write it as $\dfrac{dx}{dy}+P(y)x=Q(y)$ and use the integrating factor $e^{\int P\,dy}$.

   Step-by-Step Solution

   1. Given: $(1+y^2)+(x-e^{\tan^{-1}y})\dfrac{dy}{dx}=0$.
   2. Rearranging: $(x-e^{\tan^{-1}y})\dfrac{dy}{dx}=-(1+y^2)$, so $\dfrac{dx}{dy}=\dfrac{x-e^{\tan^{-1}y}}{-(1+y^2)}=\dfrac{e^{\tan^{-1}y}-x}{1+y^2}$.
   3. This gives the linear form: $\dfrac{dx}{dy}+\dfrac{1}{1+y^2}x=\dfrac{e^{\tan^{-1}y}}{1+y^2}$.
   4. Integrating factor: $I=e^{\int \frac{dy}{1+y^2}}=e^{\tan^{-1}y}$.
   5. Multiply through: $\dfrac{d}{dy}\left(x\,e^{\tan^{-1}y}\right)=e^{\tan^{-1}y}\cdot\dfrac{e^{\tan^{-1}y}}{1+y^2}=\dfrac{e^{2\tan^{-1}y}}{1+y^2}$.
   6. Let $t=\tan^{-1}y$, so $dt=\dfrac{dy}{1+y^2}$. Then $\int e^{2t}\,dt=\tfrac12 e^{2t}$. …

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10. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.The integrating factor of the linear differential equation $\dfrac{dy}{dx} = \dfrac{1}{4x+3y}$ is (A) $e^{-4x}$ (B) $e^{4x}$ (C) $e^{3y}$ (D) $e^{-3y}$

    ›Reveal solutionSolution

    The ODE is not linear in $y$, but writing $x$ as the dependent variable turns it into $\dfrac{dx}{dy}-4x=3y$, whose integrating factor is $e^{-4y}$ — the option carrying the exponent $-4$.

    Concept and Intuition

    A first-order linear ODE must look like $\dfrac{dy}{dx}+P(x)y=Q(x)$. Here $\dfrac{dy}{dx}=\dfrac{1}{4x+3y}$ has the unknown buried in a denominator, so it is not linear in $y$. The rescue is a change of viewpoint: nothing stops us from regarding $x$ as the function and $y$ as the independent variable. Since $\dfrac{dx}{dy}=1\big/\dfrac{dy}{dx}$, the reciprocal instantly clears the denominator and the equation becomes linear in $x$. The integrating factor is then $e^{\int P\,dy}$, with $P$ the coefficient of $x$.

    Step-by-Step Solution

    1. Invert: $\dfrac{dx}{dy}=4x+3y$.
    2. Put it in standard linear form (dependent variable $x$):

    $\frac{dx}{dy}-4x=3y,\qquad P(y)=-4,\ \ Q(y)=3y .$

    3. Integrating factor: $\text{I.F.}=e^{\int P\,dy}=e^{\int(-4)\,dy}=e^{-4y}.$ …

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11. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.The general solution of the differential equation $\dfrac{dy}{dx}+\dfrac{\sec x}{\cos x + \sin x}y=\dfrac{\cos x}{1+\tan x}$ is (A) $(\cos x + \sin x)y = \sin x + c$ (B) $(\cos x + \sin x)y = \cos x + c$ (C) $(1+\tan x)y = \cos x + c$ (D) $\sec x(\cos x + \sin x)y = \sin x + c$

    ›Reveal solutionSolution

    Recognising $P(x)=\sec x/(\cos x+\sin x)$ as $\sec^2x/(1+\tan x)$ gives the integrating factor $1+\tan x$; the equation then integrates cleanly to $\sec x(\cos x+\sin x)y=\sin x+c$.

    Concept and Intuition

    For a linear ODE $y'+P(x)y=Q(x)$, the integrating factor is $e^{\int P\,dx}$. Spotting that a messy $P(x)$ is secretly of the form $\dfrac{u'(x)}{u(x)}$ for some simple $u(x)$ (here $u=1+\tan x$) instantly gives $\int P\,dx=\log|u|$ without a hard integration.

    Step-by-Step Solution

    1. Here $P(x)=\dfrac{\sec x}{\cos x+\sin x}$. Multiply numerator and denominator by $\sec x$: $\dfrac{\sec^2x}{\sec x(\cos x+\sin x)}=\dfrac{\sec^2x}{1+\tan x}$ (since $\cos x(\cos x+\sin x)=\cos^2x(1+\tan x)$).
    2. So $P(x)=\dfrac{\sec^2x}{1+\tan x}=\dfrac{d}{dx}\log(1+\tan x)$, giving integrating factor $\mathrm{IF}=1+\tan x$.
    3. $Q(x)=\dfrac{\cos x}{1+\tan x}$, so $Q\cdot\mathrm{IF}=\dfrac{\cos x}{1+\tan x}\cdot(1+\tan x)=\cos x$.
    4. The linear-ODE solution is $y\cdot\mathrm{IF}=\int Q\cdot\mathrm{IF}\,dx+c$: $(1+\tan x)y=\displaystyle\int\cos x\,dx+c=\sin x+c$. …

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12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.Find the particular solution of the following differential equation, given that $y = 1$ when $x = 0$. $(1+x^2)\dfrac{dy}{dx} = e^{m(\text{Tan}^{-1}(x))} - y$ (A) $x\,e^{\text{Tan}^{-1}(x)} = \text{Tan}^{-1}(x) + 1$ (B) $x\,e^{\text{Tan}^{-1}(x)} = \text{Tan}^{-1}(x) - 1$ (C) $y\,e^{\text{Tan}^{-1}(x)} = \text{Tan}^{-1}(x) + 1$ (D) $y\,e^{\text{Tan}^{-1}(x)} = \text{Tan}^{-1}(x) - 1$

    ›Reveal solutionSolution

    This is a first-order linear ODE in $y$; the integrating factor $e^{\tan^{-1}x}$ turns the left side into an exact derivative, and the initial condition fixes the constant to $1$.

    Concept and Intuition

    Any equation of the form $(1+x^2)y' + y = g(x)$ is linear, since dividing by $(1+x^2)$ gives $y' + \dfrac{y}{1+x^2} = \dfrac{g(x)}{1+x^2}$, and the coefficient of $y$, $\dfrac{1}{1+x^2}$, integrates to $\tan^{-1}x$ — so the integrating factor is always $e^{\tan^{-1}x}$ for this family of equations.

    Step-by-Step Solution

    1. Rewrite: $\dfrac{dy}{dx}+\dfrac{y}{1+x^2} = \dfrac{e^{-\tan^{-1}x}}{1+x^2}$.
    2. Integrating factor: $\mu = e^{\int \frac{dx}{1+x^2}} = e^{\tan^{-1}x}$.
    3. Multiplying through: $\dfrac{d}{dx}\big[y\,e^{\tan^{-1}x}\big] = \dfrac{e^{\tan^{-1}x}\cdot e^{-\tan^{-1}x}}{1+x^2} = \dfrac{1}{1+x^2}$.
    4. Integrate both sides: $y\,e^{\tan^{-1}x} = \tan^{-1}x + C$. …

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