Q.General solution of $\frac{dy}{dx}+y\tan x=\sec x$ is:
(A) $y\sec x=\tan x+c$
(B) $y\tan x=\sec x+c$
(C) $\tan x=y\tan x+c$
(D) $x\sec x=\tan y+c$

Showing the 12 most recent of 33 on this concept.

1. 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 $y\cdot\mathrm{IF}=\int Q\cdot\mathrm{IF}\,dx+c$: $(1+\tan x)y=\displaystyle\int\cos x\,dx+c=\sin x+c$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 →
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 $u=\log(\cos x)$, $dv=\sec x\tan x\,dx\Rightarrow v=\sec x$: $\displaystyle\int\sec x\tan x\log(\cos x)dx=\sec x\log(\cos x)-\int\sec x\cdot(-\tan x)\,dx=\sec x\log(\cos x)+\sec x+C$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $\dfrac{d}{dy}(x\tan y)=\cos^3y\cdot\tan y=\cos^3y\cdot\dfrac{\sin y}{\cos y}=\cos^2y\sin y$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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: $\mu(y)=\exp\!\left[\int\left(1+\frac{\cos y-\sin y}{\cos y+\sin y}\right)dy\right] = \exp\left[y+\log|\cos y+\sin y|\right] = e^y(\cos y+\sin y)$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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: $xy=e^x+c$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 →
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 2024Set eng-2024-05-20-FN1 markMCQ
   Q.The general solution of the differential equation $(y^2+x+1)dy=(y+1)dx$ is (A) $x+2+(y+1)\log(y+1)^2=y+c$ (B) $x+2+\log(y+1)^2=\dfrac{y}{y+1}+c$ (C) $\dfrac{x}{y+1}=\log(y+1)^2+y+c$ (D) $\dfrac{x+2}{y+1}+\log(y+1)^2=y+c$

   ›Reveal solutionSolution

   This is a linear differential equation once you treat $x$ as the dependent variable and $y$ as the independent variable; solving it and simplifying the constant gives option (D).

   Concept and Intuition

   The equation $(y^2+x+1)\,dy=(y+1)\,dx$ mixes $x$ and $y$ in a way that is NOT separable and NOT linear in $y$ as a function of $x$. But if we flip our viewpoint and treat $x$ as a function of $y$, the equation becomes linear in $x$ — this is a common trick: whenever the "wrong" variable makes the equation linear, solve for that one instead.

   Step-by-Step Solution

   1. Divide by $(y+1)\,dy$:

   $\frac{dx}{dy} = \frac{y^2+x+1}{y+1} = \frac{x}{y+1} + \frac{y^2+1}{y+1}$

   2. Rearrange into standard linear form $\dfrac{dx}{dy} - \dfrac{1}{y+1}x = \dfrac{y^2+1}{y+1}$, so $P(y) = -\dfrac{1}{y+1}$, $Q(y) = \dfrac{y^2+1}{y+1}$.
   3. Integrating factor: $\mu = e^{\int P\,dy} = e^{-\log(y+1)} = \dfrac{1}{y+1}$.
   4. The solution is $x\cdot\mu = \int Q\cdot\mu\,dy$, i.e.

   $\frac{x}{y+1} = \int \frac{y^2+1}{(y+1)^2}\,dy$

   5. Substitute $u=y+1$ (so $y=u-1$, $y^2+1 = u^2-2u+2$):

   $\frac{y^2+1}{(y+1)^2} = \frac{u^2-2u+2}{u^2} = 1 - \frac{2}{u} + \frac{2}{u^2}$

   6. Integrate: $\displaystyle\int\left(1-\frac{2}{u}+\frac{2}{u^2}\right)du = u - 2\log u - \frac{2}{u} + C$, i.e.

   $\frac{x}{y+1} = (y+1) - 2\log(y+1) - \frac{2}{y+1} + C$

   7. Add $\dfrac{2}{y+1}$ to both sides: …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.The general solution of the differential equation $x\log x\,dy = (x\log x - y)\,dx$ is (A) $(x - y)\log x + x = c$ (B) $x - y = \dfrac{x}{\log x} + c$ (C) $y - x = \dfrac{x}{\log x} + c$ (D) $(y - x)\log x + x = c$

    ›Reveal solutionSolution

    This tests recognizing a first-order LINEAR differential equation in disguise and applying the standard integrating-factor method.

    Concept and Intuition

    After dividing through by $x\log x$, the equation takes the standard linear form $\dfrac{dy}{dx}+P(x)y=Q(x)$ with $P(x)=\dfrac{1}{x\log x}$. The integrating factor $e^{\int P\,dx}$ makes the left side an exact derivative $\dfrac{d}{dx}(y\cdot \text{IF})$, so the whole equation integrates directly.

    Step-by-Step Solution

    1. Divide both sides by $x\log x$: $\dfrac{dy}{dx}=1-\dfrac{y}{x\log x}\ \Rightarrow\ \dfrac{dy}{dx}+\dfrac{y}{x\log x}=1$.
    2. Integrating factor: $\mathrm{IF}=\exp\!\left(\int \dfrac{dx}{x\log x}\right)=\exp(\log(\log x))=\log x$ (since $\int\frac{dx}{x\log x}=\log(\log x)+C$).
    3. Multiply the ODE by $\log x$: $\dfrac{d}{dx}\big(y\log x\big)=\log x$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.The solution of differential equation $(x + 2y^3)\dfrac{dy}{dx} = y$ is (A) $x = y(2xy + c)$ (B) $x = y(y^2 + c)$ (C) $y = x(x^2 + c)$ (D) $xy = \dfrac{y^4}{2} + c$

    ›Reveal solutionSolution

    Flipping the roles of $x$ and $y$ turns this into a standard linear ODE in $x$; solving it gives $x = y(y^2+c)$.

    Concept and Intuition

    The equation $(x+2y^3)\dfrac{dy}{dx}=y$ is not linear in $y$, but if we invert it and treat $x$ as a function of $y$, it becomes linear in $x$ — a very common trick when the equation is "linear except the dependent/independent variables are swapped."

    Step-by-Step Solution

    1. Invert: $\dfrac{dx}{dy}=\dfrac{x+2y^3}{y}$.
    2. Rearrange: $\dfrac{dx}{dy}-\dfrac{1}{y}x = 2y^2$. This is linear in $x$ with $P(y)=-\dfrac{1}{y}$, $Q(y)=2y^2$.
    3. Integrating factor: $IF=e^{\int P\,dy}=e^{-\int dy/y}=e^{-\log y}=\dfrac{1}{y}$.
    4. Multiply through: $\dfrac{d}{dy}\!\left(\dfrac{x}{y}\right)=\dfrac{2y^2}{y}=2y$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.The general solution of the differential equation $(1+\sin^2 x)\dfrac{dy}{dx} + y\sin 2x = \cos x + \sin^2 x \cos x$ is (A) $(\sin 2x)y = \sin^2 x + c$ (B) $(1+\sin^2 x)y = \sin x - \dfrac{\sin^3 x}{3} + c$ (C) $(1+\sin^2 x)y = \sin x + \dfrac{\sin^3 x}{3} + c$ (D) $(\sin 2x)y = \sin x + \sin^2 x + c$

    ›Reveal solutionSolution

    The equation is linear in $y$ with integrating factor $1+\sin^2x$; once multiplied through, the left side collapses to $\frac{d}{dx}[(1+\sin^2x)y]$ and the right side integrates cleanly.

    Concept and Intuition

    A first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$ has integrating factor $e^{\int P\,dx}$. Here, spotting that the RHS conveniently factors to cancel the $(1+\sin^2x)$ divisor makes $Q(x)$ simplify to just $\cos x$, and $P(x)=\dfrac{\sin2x}{1+\sin^2x}$ integrates to $\log(1+\sin^2x)$ neatly.

    Step-by-Step Solution

    1. Divide by $(1+\sin^2x)$: $\dfrac{dy}{dx}+\dfrac{\sin2x}{1+\sin^2x}y=\dfrac{\cos x+\sin^2x\cos x}{1+\sin^2x}$.
    2. RHS numerator factors: $\cos x+\sin^2x\cos x=\cos x(1+\sin^2x)$, so RHS $=\cos x$ exactly. Equation becomes: $\dfrac{dy}{dx}+\dfrac{\sin2x}{1+\sin^2x}y=\cos x$.
    3. Integrating factor: $\mu=e^{\int \frac{\sin2x}{1+\sin^2x}dx}$. Since $\dfrac{d}{dx}(1+\sin^2x)=2\sin x\cos x=\sin2x$, the integral is $\log(1+\sin^2x)$, so $\mu=1+\sin^2x$.
    4. Multiply through: $\dfrac{d}{dx}\big[(1+\sin^2x)y\big]=(1+\sin^2x)\cos x=\cos x+\sin^2x\cos x$. …

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