Q.Integrating factor of the differential equation $\frac{dy}{dx}+y\tan x-\sec x=0$ is:
(A) $\cos x$
(B) $\sec x$
(C) $e^{\cos x}$
(D) $e^{\sec x}$

Showing the 12 most recent of 33 on this concept.

1. 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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2. 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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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} + y\tan x = -\tan x \log(\cos x)$ $\left(0 < x < \dfrac{\pi}{2}\right)$ (A) $\sec x = e \cdot e^{y - k\cos x}$ (B) $\sec x = e \cdot e^{y - k\sec x}$ (C) $\tan x = e \cdot e^{y - k\cos x}$ (D) $\tan x = e \cdot e^{y + k\sec x}$

   ›Reveal solutionSolution

   A first-order linear ODE with integrating factor $\sec x$; solving it and rearranging the logarithmic solution into exponential form reproduces exactly the printed option. Answer: $\sec x=e\cdot e^{y-k\cos x}$.

   Concept and Intuition

   The equation is a standard linear ODE $\frac{dy}{dx}+P(x)y=Q(x)$ with $P(x)=\tan x$. Its integrating factor is $e^{\int\tan x\,dx}=e^{-\log\cos x}=\sec x$. After solving, the answer is naturally logarithmic; the answer choices present it exponentiated, so the final algebraic step is just re-expressing $\log(\cos x)=\ldots$ as $\sec x=e\cdot e^{(\ldots)}$.

   Step-by-Step Solution

   1. Standard form: $\dfrac{dy}{dx}+y\tan x=-\tan x\log(\cos x)$, $P(x)=\tan x$, $Q(x)=-\tan x\log(\cos x)$.
   2. Integrating factor: $\mathrm{IF}=e^{\int\tan x\,dx}=e^{-\log(\cos x)}=\dfrac{1}{\cos x}=\sec x$.
   3. Solution: $y\sec x=\displaystyle\int \sec x\cdot\left(-\tan x\log(\cos x)\right)dx+k=-\int\sec x\tan x\log(\cos x)\,dx+k$.
   4. Integrate by parts with $u=\log(\cos x)$, $dv=\sec x\tan x\,dx\Rightarrow v=\sec x$: $\displaystyle\int\sec x\tan x\log(\cos x)dx=\sec x\log(\cos x)-\int\sec x\cdot(-\tan x)\,dx=\sec x\log(\cos x)+\sec x+C$. …

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4. 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 $y\cdot\dfrac{1+\sin x}{\cos x}=x-\cos x+C$. …

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5. 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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6. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The general solution of the differential equation $(1 + \tan y)(dx - dy) + 2x\,dy = 0$ is (A) $e^x(y\cos x + \sin x) + \sin x = c$ (B) $e^x(y\cos x + y\sin x - \sin x) + \cos x = 0$ (C) $e^y(x\cos y + x\sin y - \sin y) = c$ (D) $e^y(x\cos y + x\sin y + \sin y) = c$

   ›Reveal solutionSolution

   This is a first-order linear ODE in $x$ as a function of $y$; finding the right integrating factor is the crux. Answer: option (C).

   Concept and Intuition

   Grouping the $dx$ terms and $dy$ terms shows this is linear in $x$ (treating $y$ as the independent variable), of the form $\frac{dx}{dy}+P(y)x=Q(y)$. The integrating factor $e^{\int P\,dy}$ simplifies neatly once we split $2\cos y/(\cos y+\sin y)$ using the identity for a sum/difference of sine and cosine.

   Step-by-Step Solution

   1. Expand: $(1+\tan y)dx - (1+\tan y)dy + 2x\,dy = 0 \Rightarrow (1+\tan y)dx + \left[2x-(1+\tan y)\right]dy=0$.
   2. Divide by $(1+\tan y)\,dy$: $\dfrac{dx}{dy} + \dfrac{2x}{1+\tan y} = 1$.
   3. Write $\dfrac{2}{1+\tan y} = \dfrac{2\cos y}{\cos y+\sin y}$. Using $2\cos y=(\cos y+\sin y)+(\cos y-\sin y)$: $\frac{2\cos y}{\cos y+\sin y}=1+\frac{\cos y-\sin y}{\cos y+\sin y}.$
   4. Integrating factor: $\mu(y)=\exp\!\left[\int\left(1+\frac{\cos y-\sin y}{\cos y+\sin y}\right)dy\right] = \exp\left[y+\log|\cos y+\sin y|\right] = e^y(\cos y+\sin y)$. …

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7. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The general solution of the differential equation $(\sin y\cos^2 y - x\sec^2 y)\,dy=(\tan y)\,dx$ is (A) $\tan y=3x\cos^3 y+c$ (B) $x(\sec y+\tan y)=\cos^2 y+c$ (C) $y\sin y=x^2\cos^2 y+c$ (D) $3x\tan y+\cos^3 y=c$

   ›Reveal solutionSolution

   Treating $x$ as the dependent variable turns this into a linear ODE with integrating factor $\tan y$, giving $3x\tan y+\cos^3y=c$.

   Concept and Intuition

   When an ODE is not linear in $y$ but becomes linear if we treat $x$ as a function of $y$ instead, switching the roles of dependent/independent variable is the key move — here $dx/dy$ appears linearly in $x$, so it is a standard first-order linear equation solvable via an integrating factor.

   Step-by-Step Solution

   1. Given: $(\sin y\cos^2y - x\sec^2y)\,dy=\tan y\,dx$. Solve for $dx/dy$: $\dfrac{dx}{dy}=\dfrac{\sin y\cos^2y-x\sec^2y}{\tan y}$.
   2. Split: $\dfrac{\sin y\cos^2y}{\tan y}=\sin y\cos^2y\cdot\dfrac{\cos y}{\sin y}=\cos^3y$, and $\dfrac{\sec^2y}{\tan y}=\dfrac{1}{\cos^2y}\cdot\dfrac{\cos y}{\sin y}=\dfrac{1}{\sin y\cos y}$.
   3. So $\dfrac{dx}{dy}=\cos^3y-\dfrac{x}{\sin y\cos y}$, i.e. $\dfrac{dx}{dy}+\dfrac{x}{\sin y\cos y}=\cos^3y$ — linear in $x$.
   4. Integrating factor: $\mu=\exp\left(\displaystyle\int\dfrac{dy}{\sin y\cos y}\right)$. Since $\dfrac{d}{dy}\log(\tan y)=\dfrac{\sec^2y}{\tan y}=\dfrac{1}{\sin y\cos y}$, we get $\mu=\tan y$.
   5. Then $\dfrac{d}{dy}(x\tan y)=\cos^3y\cdot\tan y=\cos^3y\cdot\dfrac{\sin y}{\cos y}=\cos^2y\sin y$. …

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8. 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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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 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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11. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.The general solution of the equation $\dfrac{dy}{dx}+\dfrac{1}{x}y=\dfrac{1}{x}e^x$ is (A) $y=xe^x+c$ (B) $y=xe^x+ce^{-x}$ (C) $y=\dfrac{e^x+c}{x}$ (D) $y=\dfrac{e^{-x}+cx}{x}$

    ›Reveal solutionSolution

    This is a standard linear first-order ODE solved via an integrating factor; the answer is (C).

    Concept and Intuition

    An equation of the form $\dfrac{dy}{dx}+P(x)y=Q(x)$ is solved by multiplying through by the integrating factor $\mu=e^{\int P\,dx}$, which makes the left side a perfect derivative $\dfrac{d}{dx}(\mu y)$.

    Step-by-Step Solution

    1. Here $P(x)=\dfrac1x$, $Q(x)=\dfrac1x e^x$.
    2. Integrating factor: $\mu=e^{\int \frac1x dx}=e^{\log x}=x$.
    3. Multiply the ODE by $x$: $x\dfrac{dy}{dx}+y=e^x$, i.e. $\dfrac{d}{dx}(xy)=e^x$.
    4. Integrate both sides: $xy=e^x+c$. …

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12. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.By multiplying with $e^{\int P\,dx}$ on both sides of the equation $\dfrac{dy}{dx} + P(x)y = Q(x)$, the left side of the equation takes the form $\dfrac{d}{dx}(y\,f(x))$, then $f(x) =$ (A) $\displaystyle\int y\,e^{\int P\,dx}\,dx$ (B) $y\,P(x)$ (C) $e^{\int P\,dx}$ (D) $P(x)\,e^{\int P\,dx}$

    ›Reveal solutionSolution

    This is the definition-check behind the standard integrating-factor method for first-order linear ODEs: the multiplier that turns the left side into an exact derivative $\frac{d}{dx}(y\cdot(\text{something}))$ is, by construction, the integrating factor itself.

    Concept and Intuition

    The whole point of the integrating factor $\mu(x)=e^{\int P\,dx}$ is that $\dfrac{d\mu}{dx}=P(x)\mu(x)$ (by the fundamental theorem of calculus applied to the exponent). This special property is exactly what's needed to make $\mu\dfrac{dy}{dx}+\mu P y$ collapse into a single product-rule derivative.

    Step-by-Step Solution

    1. Start with $\dfrac{dy}{dx}+P(x)y=Q(x)$ and multiply both sides by $\mu(x)=e^{\int P\,dx}$:

    $\mu\frac{dy}{dx}+\mu P(x)y=\mu Q(x).$

    2. Compute $\dfrac{d\mu}{dx}$: since $\mu=e^{\int P\,dx}$, by the chain rule $\dfrac{d\mu}{dx}=P(x)\cdot e^{\int P\,dx}=P(x)\mu$. …

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