Q.The general solution of the differential equation $\frac{y\, dx - x\, dy}{y} = 0$ is (A) $xy = C$ (B) $x = Cy^2$ (C) $y = Cx$ (D) $y = Cx^2$

Showing the 12 most recent of 33 on this concept.

1. 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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2. 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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3. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.The solution of the differential equation $e^xy\,dx + e^x\,dy + x\,dx = 0$ is (A) $e^x+yx^2=c$ (B) $2ye^x+x^2=c$ (C) $ye^x+x^2e^y=c$ (D) $e^x+xe^y=c$

   ›Reveal solutionSolution

   The first two terms of the equation are exactly $d(ye^x)$; separating out the remaining $x\,dx$ term and integrating directly gives the solution.

   Concept and Intuition

   Many differential equations that look complicated are secretly "exact" — the left side is the total differential of some simple combination of $x$ and $y$. Spotting the pattern $u\,dv+v\,du=d(uv)$ (here with $u=y,\ v=e^x$) turns an equation that looks like it needs an integrating factor into a one-line integration.

   Step-by-Step Solution

   1. Given: $e^xy\,dx+e^x\,dy+x\,dx=0$.
   2. Recall $d(ye^x)=y\,d(e^x)+e^x\,dy=ye^x\,dx+e^x\,dy$ — exactly the first two terms of the given equation.
   3. So the equation becomes $d(ye^x)+x\,dx=0$, i.e. $d(ye^x)=-x\,dx$.
   4. Integrate both sides: $ye^x=-\dfrac{x^2}{2}+C$.
   5. Multiply through by 2: $2ye^x=-x^2+2C$, i.e. $2ye^x+x^2=c$ (writing $c=2C$).

   Common Mistakes …

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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 $\dfrac{d}{dy}(x\tan y)=\cos^3y\cdot\tan y=\cos^3y\cdot\dfrac{\sin y}{\cos y}=\cos^2y\sin y$. …

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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: $\mu(y)=\exp\!\left[\int\left(1+\frac{\cos y-\sin y}{\cos y+\sin y}\right)dy\right] = \exp\left[y+\log|\cos y+\sin y|\right] = e^y(\cos y+\sin y)$. …

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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 $\int e^{2t}\,dt=\tfrac12 e^{2t}$. …

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7. 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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8. 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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9. 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: $\dfrac{d}{dy}\!\left(\dfrac{x}{y}\right)=\dfrac{2y^2}{y}=2y$. …

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10. 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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11. 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 $\log x$: $\dfrac{d}{dx}\big(y\log x\big)=\log x$. …

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12. 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: …

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