Q.Solve the differential equation $\left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\right) \frac{dx}{dy} = 1\ (x \neq 0)$.

Showing the 12 most recent of 33 on this concept.

1. 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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2. 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: $\mu = e^{\int 2x\,dx}=e^{x^2}$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx}+\dfrac{y}{x}=x^2$ is (A) $y=\dfrac{1}{3}x^3+\dfrac{c}{x}$ (B) $y=\dfrac{1}{4}x^4+cx$ (C) $y=\dfrac{1}{4}x^3+c$ (D) $y=\dfrac{1}{4}x^3+cx^{-1}$

   ›Reveal solutionSolution

   A standard first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$; the integrating factor $x$ makes the left side an exact derivative, giving the general solution $y=\dfrac14x^3+\dfrac{c}{x}$.

   Concept and Intuition

   For a linear equation $\dfrac{dy}{dx}+P(x)y=Q(x)$, multiplying both sides by the integrating factor $\mu(x)=e^{\int P\,dx}$ turns the left side into the exact derivative $\dfrac{d}{dx}\big(\mu y\big)$, which can then be integrated directly.

   Step-by-Step Solution

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

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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: $-\dfrac{1}{u}=-\log y+c_1$. …

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5. 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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6. 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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7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The substitution required to reduce the differential equation $\dfrac{dy}{dx} + \sin y \cos y \sin x = \sin 2x \cos^2 y$ to a linear differential equation in $z$ is (A) $z = \tan x$ (B) $z = \sin 2y$ (C) $z = \cos y$ (D) $z = \tan y$

   ›Reveal solutionSolution

   Dividing by $\cos^2y$ and substituting $z=\tan y$ converts the equation into a linear first-order ODE in $z$.

   Concept and Intuition

   Equations with $\sin y\cos y$ and $\cos^2y$ terms often become linear after dividing through by $\cos^2y$, because $\dfrac{d}{dx}(\tan y)=\sec^2y\,\dfrac{dy}{dx}$ naturally appears in the divided equation.

   Step-by-Step Solution

   1. Given: $\dfrac{dy}{dx}+\sin y\cos y\sin x=\sin 2x\cos^2y=2\sin x\cos x\cos^2y$.
   2. Divide throughout by $\cos^2y$:

   $\sec^2y\frac{dy}{dx}+\tan y\sin x=2\sin x\cos x$

   3. Let $z=\tan y \Rightarrow \dfrac{dz}{dx}=\sec^2y\dfrac{dy}{dx}$. Substituting: $\frac{dz}{dx}+z\sin x=2\sin x\cos x$ …

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8. 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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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} + xy = 4x - 2y + 8$ is (A) $y = 4 - ce^{-\frac{(x+2)^2}{2}}$ (B) $y = 8 + ce^{\frac{-x^2}{2} - 2x}$ (C) $y = ce^{-(x+2)^2} + x$ (D) $y + 2x = ce^{-\frac{x}{2} - 2x}$

   ›Reveal solutionSolution

   Regroup the equation into standard linear form $y'+(x+2)y=4(x+2)$, solve with integrating factor $e^{x^2/2+2x}$, and complete the square in the exponent. Answer: $y=4-ce^{-(x+2)^2/2}$.

   Concept and Intuition

   The given equation $\dfrac{dy}{dx}+xy = 4x-2y+8$ looks like it has an $x$-dependent coefficient and a constant-coefficient term mixed together, but moving the $-2y$ across makes the coefficient of $y$ become $(x+2)$, and simultaneously the right side becomes $4(x+2)$ — a clean multiple of the same linear factor. This is the key algebraic regrouping that turns it into a standard first-order linear ODE.

   Step-by-Step Solution

   1. Start: $\dfrac{dy}{dx}+xy = 4x-2y+8$.
   2. Move $-2y$ to the left: $\dfrac{dy}{dx}+xy+2y = 4x+8 \Rightarrow \dfrac{dy}{dx}+(x+2)y = 4(x+2)$.
   3. This is linear: $\dfrac{dy}{dx}+P(x)y=Q(x)$ with $P(x)=x+2$, $Q(x)=4(x+2)$.
   4. Integrating factor: $\mu(x)=e^{\int(x+2)dx}=e^{\frac{x^2}{2}+2x}$.
   5. Note $\dfrac{d\mu}{dx}=(x+2)\mu$, so $4(x+2)\mu = 4\dfrac{d\mu}{dx}$, and

   $\frac{d}{dx}(y\mu) = Q(x)\mu = 4\frac{d\mu}{dx} \implies y\mu = 4\mu + C.$

   6. Hence $y = 4 + Ce^{-\left(\frac{x^2}{2}+2x\right)}$.
   7. Complete the square: $\dfrac{x^2}{2}+2x = \dfrac{x^2+4x}{2} = \dfrac{(x+2)^2-4}{2} = \dfrac{(x+2)^2}{2}-2$. …

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10. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.The general solution of the differential equation $(x + 2y^3)\dfrac{dy}{dx} - y = 0$, $y > 0$ is (A) $y = x^3 + cy$ (B) $x = y^3 + cy$ (C) $y(1 - xy) = cx$ (D) $x(1 - xy) = cy$

    ›Reveal solutionSolution

    Treat $x$ as the dependent variable and $y$ as independent — the equation is linear in $x$ once rearranged. Answer: $x=y^3+cy$.

    Concept and Intuition

    The equation $(x+2y^3)\dfrac{dy}{dx}-y=0$ is not linear in $y$ (because of the $y^3$ term), but if we instead regard $x$ as a function of $y$, it becomes linear in $x$. This is a standard trick: whenever an ODE is linear in one variable when viewed "the other way around," solve it as $\dfrac{dx}{dy}+P(y)x=Q(y)$ instead of $\dfrac{dy}{dx}+P(x)y=Q(x)$.

    Step-by-Step Solution

    1. Given: $(x+2y^3)\dfrac{dy}{dx} = y \implies \dfrac{dx}{dy} = \dfrac{x+2y^3}{y}$ (inverting the derivative, valid since $y>0$).
    2. Expand: $\dfrac{dx}{dy} = \dfrac{x}{y} + 2y^2 \implies \dfrac{dx}{dy} - \dfrac{1}{y}x = 2y^2$.
    3. This is linear in $x$ with $P(y)=-\dfrac1y$, $Q(y)=2y^2$.
    4. Integrating factor: $\mu(y) = e^{\int -\frac1y dy} = e^{-\log y} = \dfrac1y$.
    5. Then $\dfrac{d}{dy}\!\left(\dfrac{x}{y}\right) = \dfrac1y\cdot 2y^2 = 2y$.
    6. Integrate: $\dfrac{x}{y} = \displaystyle\int 2y\,dy = y^2 + c$. …

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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 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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