Q.Integrating factor of the differential equation $(1-x^2)\frac{dy}{dx}-xy=1$ is:
(A) $-x$
(B) $\frac{x}{1+x^2}$
(C) $\sqrt{1-x^2}$
(D) $\frac{1}{2}\log(1-x^2)$

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 $u=x^2+1$). …

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

   🔒 Start your 14-day free trial to see the full solution →
3. 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. $\dfrac{d}{dx}\!\left(\dfrac{v}{x}\right)=\dfrac{1}{x} \Rightarrow \dfrac{v}{x}=\log x+C \Rightarrow v=x(\log x+C)$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The general solution of the differential equation $y\,dx + (x+x^2y)\,dy = 0$ is (A) $\dfrac{1}{xy}+\log y = c$ (B) $-\dfrac{1}{xy}+\log y = c$ (C) $x-\dfrac{1}{xy}=c$ (D) $\log y = cx^2$

   ›Reveal solutionSolution

   This tests recognizing an exact-differential grouping ($y\,dx+x\,dy=d(xy)$) to reduce the equation to a separable one; the answer is (B).

   Concept and Intuition

   The equation $y\,dx+(x+x^2y)\,dy=0$ looks messy until you notice that $y\,dx+x\,dy$ is exactly $d(xy)$. Recognizing hidden exact-differential combinations (like $d(xy)$, $d(x/y)$, $d(x^2+y^2)$) is often the fastest route through an ODE that doesn't look separable or linear at first glance.

   Step-by-Step Solution

   1. Rewrite: $y\,dx+x\,dy+x^2y\,dy=0$.
   2. Since $d(xy)=x\,dy+y\,dx$, this becomes $d(xy)+x^2y\,dy=0$.
   3. Let $u=xy$, so $x=u/y$. Then $x^2y = \dfrac{u^2}{y^2}\cdot y=\dfrac{u^2}{y}$.
   4. Substituting: $du+\dfrac{u^2}{y}dy=0 \Rightarrow \dfrac{du}{u^2}=-\dfrac{dy}{y}$.
   5. Integrate both sides: $-\dfrac{1}{u}=-\log y+c_1$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $\dfrac{d}{dx}\left[y(x^2+1)^2\right]$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 $\int e^{2t}\,dt=\tfrac12 e^{2t}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $w=1-x^2$, $dw=-2x\,dx$): $\int\dfrac{x\,dx}{(1-x^2)^{3/2}}=\dfrac{1}{\sqrt{1-x^2}}+C$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} = x^3\cos^2 y - x\sin 2y$ is (A) $\tan y = \dfrac{1}{2}(x^2 + 1) + e^{-x^2}$ (B) $\tan y = \dfrac{1}{2}(x^2 - 1) + Ce^{-x^2}$ (C) $\tan y = \dfrac{1}{2}(x^2 - 1) + Ce^{x^2}$ (D) $\tan y = \dfrac{1}{2}(x^2 + 1) + Ce^{x^2}$

   ›Reveal solutionSolution

   This tests converting a nonlinear-looking ODE in $y$ into a linear first-order ODE via the substitution $t=\tan y$, then solving with an integrating factor. The answer is $\tan y = \tfrac12(x^2-1)+Ce^{-x^2}$.

   Concept and Intuition

   The presence of $\cos^2 y$ and $\sin 2y=2\sin y\cos y$ is a strong hint to divide through by $\cos^2 y$: this turns every $y$-term into a function of $\tan y$, because $\sin 2y/\cos^2y = 2\tan y$. Substituting $t=\tan y$ then reduces the equation to the standard linear form $\dfrac{dt}{dx}+P(x)t=Q(x)$, solvable by an integrating factor.

   Step-by-Step Solution

   1. Start with $\dfrac{dy}{dx}=x^3\cos^2y - x\sin2y = x^3\cos^2 y - 2x\sin y\cos y$.
   2. Divide both sides by $\cos^2 y$: $\sec^2y\dfrac{dy}{dx} = x^3 - 2x\tan y$.
   3. Let $t=\tan y$, so $\dfrac{dt}{dx}=\sec^2y\dfrac{dy}{dx}$. The equation becomes $\dfrac{dt}{dx}+2xt = x^3$ — linear in $t$.
   4. Integrating factor: $\mu = e^{\int 2x\,dx}=e^{x^2}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $Z=f(y)=\tfrac12(x^2-1)+Ce^{-x^2}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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: $\dfrac{d}{dy}\log(\log y)=\dfrac{1}{\log y}\cdot\dfrac{d}{dy}(\log y)=\dfrac{1}{\log y}\cdot\dfrac1y=\dfrac{1}{y\log y}$. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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: $y\,e^{\tan^{-1}x} = \tan^{-1}x + C$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.The general solution of the differential equation $y + \cos x\left(\frac{dy}{dx}\right) - \cos^2 x = 0$ is (A) $(\sec x + \tan x)y = x + \cos x + c$ (B) $(1+\cos x)y = (x+c)\cos x - \cos^2 x$ (C) $(1+\sin x)y = (x+c)\cos x - \cos^2 x$ (D) $(\sec x + \tan x)y = x - \sin x + c$

    ›Reveal solutionSolution

    A first-order linear ODE in $y$; using the integrating factor $\sec x+\tan x$ and rewriting it as $(1+\sin x)/\cos x$ yields option (C).

    Concept and Intuition

    After dividing by $\cos x$, the equation becomes linear in $y$ with integrating factor $e^{\int \sec x\,dx}=\sec x+\tan x$ (a standard integral). Since $\sec x+\tan x=\frac{1+\sin x}{\cos x}$, the solution can be rewritten multiplying through by $\cos x$, which is exactly the form the answer choices use.

    Step-by-Step Solution

    1. Start with $y+\cos x\,\dfrac{dy}{dx}-\cos^2 x=0$. Divide by $\cos x$: $\dfrac{dy}{dx}+y\sec x=\cos x$.
    2. This is linear: $P(x)=\sec x$, $Q(x)=\cos x$. Integrating factor $\mu=e^{\int \sec x\,dx}=e^{\log|\sec x+\tan x|}=\sec x+\tan x$.
    3. The solution is $y\cdot\mu=\int Q\cdot\mu\,dx$: $y(\sec x+\tan x)=\int \cos x(\sec x+\tan x)\,dx=\int(1+\sin x)\,dx=x-\cos x+C.$
    4. Now write $\sec x+\tan x=\dfrac{1+\sin x}{\cos x}$, so $y\cdot\dfrac{1+\sin x}{\cos x}=x-\cos x+C$. …

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