Q.Solve the following differential equation: $y \ dx + (x - y^2) \ dy = 0$

Showing the 12 most recent of 33 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.The differential equation $y^2\, dx + (3xy - 1)\, dy = 0$ is (A) linear in y (B) not a linear equation (C) a homogenous equation (D) linear in x

   ›Reveal solutionSolution

   Treating $x$ as the dependent variable (function of $y$) puts the equation into the standard linear-ODE form.

   Concept and Intuition

   An equation is "linear in $x$" if, when we treat $x$ as the unknown function of $y$, it can be written as $\dfrac{dx}{dy}+P(y)x=Q(y)$ — i.e. $x$ and $\dfrac{dx}{dy}$ appear only to the first power, with coefficients depending on $y$ alone.

   Step-by-Step Solution

   1. Given: $y^2\,dx+(3xy-1)\,dy=0$.
   2. Divide by $dy$: $y^2\dfrac{dx}{dy}+3xy-1=0 \Rightarrow y^2\dfrac{dx}{dy}=1-3xy$.
   3. Divide by $y^2$: $\dfrac{dx}{dy}=\dfrac{1}{y^2}-\dfrac{3x}{y}$.
   4. Rearrange: $\dfrac{dx}{dy}+\dfrac{3}{y}x=\dfrac{1}{y^2}$. …

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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: $\mu = e^{\int 2x\,dx}=e^{x^2}$. …

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

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12. 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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