Q.The general solution of $\frac{dy}{dx}=2x\,e^{x^2-y}$ is:
(A) $e^{x^2-y}=c$
(B) $e^{-y}+e^{x^2}=c$
(C) $e^y=e^{x^2}+c$
(D) $e^{x^2}+y=c$

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The general solution of the differential equation $\cos(x+y)dy = dx$ is (A) $y = \tan\left(\frac{x+y}{2}\right) + c$ (B) $y = x\sec\left(\frac{y}{x}\right) + c$ (C) $y = -\mathrm{Cos}^{-1}\left(\frac{y}{x}\right) + c$ (D) $y = \tan(x+y) + c$

   ›Reveal solutionSolution

   This tests solving a differential equation by a substitution $u=x+y$ that turns it into a separable equation. The answer is (A).

   Concept and Intuition

   When an ODE contains only the combination $x+y$ (not $x$ and $y$ separately) inside a function, the standard trick is to substitute $u=x+y$. This converts the equation into one relating $u$ and one variable only, which is often separable even though the original wasn't in $x,y$ directly.

   Step-by-Step Solution

   1. The equation is $\cos(x+y)\,dy=dx$, i.e. $\dfrac{dx}{dy}=\cos(x+y)$.
   2. Let $u=x+y$. Then $\dfrac{du}{dy}=\dfrac{dx}{dy}+1$, so $\dfrac{dx}{dy}=\dfrac{du}{dy}-1$.
   3. Substituting: $\dfrac{du}{dy}-1=\cos u \Rightarrow \dfrac{du}{dy}=1+\cos u=2\cos^2\left(\dfrac{u}{2}\right)$.
   4. Separate variables: $\dfrac{du}{2\cos^2(u/2)}=dy \Rightarrow \dfrac{1}{2}\sec^2\left(\dfrac{u}{2}\right)du=dy$.
   5. Integrate: $\dfrac{1}{2}\cdot 2\tan\left(\dfrac{u}{2}\right)=y+c \Rightarrow \tan\left(\dfrac{u}{2}\right)=y+c$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The general solution of the differential equation $(4xy^2 - 2xy + 2x^2y^2 - x^2y)\, dx = dy$ is $x^2 + \dfrac{x^3}{3} =$ (A) $\log \dfrac{c(2y - 1)}{y}$ (B) $\log \dfrac{c|2y - 1|}{\sqrt{y}}$ (C) $\log c(2y^2 - y)$ (D) $\log \dfrac{c(2y - 1)}{y^2}$

   ›Reveal solutionSolution

   This is a separable (disguised) equation once the RHS is factored; separating and integrating both sides gives $x^2+\frac{x^3}{3}=\log\dfrac{c(2y-1)}{y}$.

   Concept and Intuition

   Many "ugly" first-order equations are separable after factoring out a common structure. Here $y(2y-1)$ appears twice on the right, once multiplied by $2x$ and once by $x^2$, so the whole RHS factors as $(2y-1)y\cdot x(x+2)$ — turning a messy quartic-looking expression into a clean separable form.

   Step-by-Step Solution

   1. $(4xy^2-2xy+2x^2y^2-x^2y)dx=dy \Rightarrow \dfrac{dy}{dx}=4xy^2-2xy+2x^2y^2-x^2y$.
   2. Group terms with $x$ and $x^2$ separately: $2xy(2y-1)+x^2y(2y-1)$.
   3. Factor the common $(2y-1)y$: $=(2y-1)y\,(2x+x^2)=(2y-1)y\cdot x(x+2)$.
   4. Separate variables: $\dfrac{dy}{y(2y-1)}=x(x+2)\,dx=(x^2+2x)\,dx$.
   5. Integrate the RHS: $\int (x^2+2x)dx=\dfrac{x^3}{3}+x^2$ — exactly the LHS given in the problem, confirming the factoring.
   6. Integrate the LHS by partial fractions: $\dfrac{1}{y(2y-1)}=\dfrac{A}{y}+\dfrac{B}{2y-1}$. Solving, $1=A(2y-1)+By$; at $y=0$, $A=-1$; at $y=\tfrac12$, $B=2$.
   7. So $\int\left(-\dfrac1y+\dfrac{2}{2y-1}\right)dy=-\ln|y|+\ln|2y-1|=\ln\left|\dfrac{2y-1}{y}\right|$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} = x^2 y(x + y + xy + 1)$ is $y =$ (A) $A e^{\frac{x^3}{3}} \cdot e^{\frac{x^4}{4}}(1+y)$ (B) $(y+1)\, A e^{\frac{x^3}{3}} \cdot e^{\frac{-x^4}{4}}$ (C) $(y+1) + A e^{\frac{x^3}{3}} \cdot e^{\frac{x^4}{4}}$ (D) $\dfrac{A e^{\frac{x^3}{3}}(y+1)}{e^{\frac{x^4}{4}}}$

   ›Reveal solutionSolution

   Factoring $x+y+xy+1=(1+x)(1+y)$ makes the ODE separable; integrating both sides and exponentiating gives $y=Ae^{x^3/3}e^{x^4/4}(1+y)$.

   Concept and Intuition

   Recognizing an algebraic factorization inside a differential equation is often the key to separating variables. Here $x+y+xy+1$ factors neatly as $(1+x)(1+y)$, splitting the right side into a pure function of $x$ times a pure function of $y$.

   Step-by-Step Solution

   1. Factor: $x+y+xy+1=(1+x)+y(1+x)=(1+x)(1+y)$.
   2. Rewrite the ODE:

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

   3. Separate variables:

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

   4. Partial fractions on the left: $\dfrac{1}{y(1+y)}=\dfrac1y-\dfrac{1}{1+y}$, so

   $\int\left(\frac1y-\frac{1}{1+y}\right)dy=\log|y|-\log|1+y|=\log\left|\frac{y}{1+y}\right|$

   5. Right side: $\int x^2(1+x)dx=\int(x^2+x^3)dx=\dfrac{x^3}{3}+\dfrac{x^4}{4}$.
   6. So $\log\left|\dfrac{y}{1+y}\right|=\dfrac{x^3}{3}+\dfrac{x^4}{4}+C_1$, giving …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The general solution of the differential equation $\sec(x-y+1)\,dy=dx$ is (A) $x+\cot\left(\dfrac{x-y+1}{2}\right)=c$ (B) $x+\cot(x-y+1)=c$ (C) $x-\cot\left(\dfrac{x-y+1}{2}\right)=c$ (D) $x-\cot(x-y+1)=c$

   ›Reveal solutionSolution

   Substituting $v=x-y+1$ turns this into a separable differential equation whose solution is $x+\cot\left(\frac{x-y+1}{2}\right)=c$.

   Concept and Intuition

   The equation $\sec(x-y+1)\,dy=dx$ only involves $x$ and $y$ through the combination $x-y+1$, which is the standard cue to substitute a single variable $v$ for that combination, reducing the PDE-looking equation to a simple separable ODE in $v$ and $x$.

   Step-by-Step Solution

   1. Let $v=x-y+1$, so $\dfrac{dv}{dx}=1-\dfrac{dy}{dx}$.
   2. From $\sec v\,dy=dx$: $\dfrac{dy}{dx}=\dfrac{1}{\sec v}=\cos v$.
   3. So $\dfrac{dv}{dx}=1-\cos v$.
   4. Separate variables: $\dfrac{dv}{1-\cos v}=dx$.
   5. Using $1-\cos v=2\sin^2(v/2)$: $\dfrac{1}{2}\csc^2\left(\dfrac{v}{2}\right)dv=dx$.
   6. Integrate: $\dfrac12\cdot\left[-2\cot\left(\dfrac{v}{2}\right)\right]=x+C \Rightarrow -\cot\left(\dfrac{v}{2}\right)=x+C$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.The general solution of the differential equation $xy(y+2)dy + (y^3-1)dx = 0$ is (A) $\log|x+2y| + \dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{y-x}{\sqrt{3}x}\right) = c$ (B) $\log|2x-y| + \dfrac{2}{3}\tan^{-1}\left(\dfrac{x-y}{\sqrt{3}x}\right) = c$ (C) $\log|xy-x| + \dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right) = c$ (D) $\log|x+y| + \dfrac{2}{3}\tan^{-1}\left(\dfrac{x-2y}{\sqrt{3}x}\right) = c$

   ›Reveal solutionSolution

   Since $x$ appears only linearly, treat $x$ as the dependent variable — the equation separates in $x$ and $y$, and partial fractions on the $y$-side give a log term plus an arctan term.

   Concept and Intuition

   When a first-order ODE is linear/separable in $x$ (i.e. $x$ appears only to the first power, multiplying everything), it's often easier to treat $x=x(y)$ rather than $y=y(x)$. Here that turns the whole equation into a simple separable one: $dx/x = (\text{function of }y)\,dy$.

   Step-by-Step Solution

   1. Rewrite: $xy(y+2)\,dy=-(y^3-1)\,dx \Rightarrow \dfrac{dx}{x}=-\dfrac{y(y+2)}{y^3-1}\,dy$.
   2. Factor $y^3-1=(y-1)(y^2+y+1)$.
   3. Partial fractions: $\dfrac{y^2+2y}{(y-1)(y^2+y+1)}=\dfrac{A}{y-1}+\dfrac{By+C}{y^2+y+1}$. Solving: $A=1,\ B=0,\ C=1$, so $\dfrac{y(y+2)}{y^3-1}=\dfrac1{y-1}+\dfrac1{y^2+y+1}$.
   4. So $\dfrac{dx}{x}=-\left[\dfrac1{y-1}+\dfrac1{y^2+y+1}\right]dy$.
   5. Integrate: $\ln|x|=-\ln|y-1|-\displaystyle\int\dfrac{dy}{y^2+y+1}+C$.
   6. Complete the square: $y^2+y+1=(y+\tfrac12)^2+\tfrac34$, so $\int\dfrac{dy}{y^2+y+1}=\dfrac2{\sqrt3}\tan^{-1}\!\left(\dfrac{2y+1}{\sqrt3}\right)$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The general solution of the differential equation $(2x - y)^2\,dy - 2(2x - y)^2\,dx - 2\,dx = 0$ is (A) $\log(2x - y) = 2x + c$ (B) $(2x - y)^3 + 4y = c$ (C) $(2x - y)^3 + 6x = c$ (D) $\log(2x - y) = 2y + c$

   ›Reveal solutionSolution

   This tests the substitution $v=ax+by$ for a differential equation whose right side depends only on the combination $2x-y$, turning it into a separable equation in $v$ and $x$.

   Concept and Intuition

   Whenever an ODE's coefficients depend only on a linear combination like $2x-y$ (not on $x,y$ separately), setting $v=2x-y$ collapses two variables into one, because $\dfrac{dv}{dx}=2-\dfrac{dy}{dx}$ lets us replace $\dfrac{dy}{dx}$ everywhere and separate variables in $v$ alone.

   Step-by-Step Solution

   1. Divide the given equation by $(2x-y)^2\,dx$: $\dfrac{dy}{dx}=2+\dfrac{2}{(2x-y)^2}$.
   2. Let $v=2x-y$, so $\dfrac{dv}{dx}=2-\dfrac{dy}{dx}$, i.e. $\dfrac{dy}{dx}=2-\dfrac{dv}{dx}$.
   3. Substitute: $2-\dfrac{dv}{dx}=2+\dfrac{2}{v^2}\ \Rightarrow\ -\dfrac{dv}{dx}=\dfrac{2}{v^2}\ \Rightarrow\ \dfrac{dv}{dx}=-\dfrac{2}{v^2}$.
   4. Separate: $v^2\,dv=-2\,dx$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.The general solution of the differential equation $\cos(x + y)\, dy = dx$ is (A) $y = \tan\left(\dfrac{x + y}{2}\right) + c$ (B) $y = \sec\left(\dfrac{x + y}{2}\right) + c$ (C) $y = x\sec\left(\dfrac{y}{x}\right) + c$ (D) $y = -\cos^{-1}\left(\dfrac{y}{x}\right) + c$

   ›Reveal solutionSolution

   Substituting $t=x+y$ reduces $\cos(x+y)\,dy=dx$ to a separable equation in $t,x$, giving $y=\tan\!\left(\frac{x+y}2\right)+c$.

   Concept and Intuition

   Whenever a differential equation depends on $x$ and $y$ only through the combination $x+y$ (never $x$ and $y$ separately), substituting $t=x+y$ collapses two variables into one, turning the ODE into a simple separable equation in $t$ and $x$ (or $t$ and $y$).

   Step-by-Step Solution

   1. From $\cos(x+y)\,dy=dx$: $\dfrac{dy}{dx}=\sec(x+y)$.
   2. Let $t=x+y$, so $\dfrac{dt}{dx}=1+\dfrac{dy}{dx}=1+\sec t$.
   3. Separate: $dx=\dfrac{dt}{1+\sec t}=\dfrac{\cos t\,dt}{\cos t+1}$.
   4. Use $1+\cos t=2\cos^2(t/2)$ and $\cos t = 2\cos^2(t/2)-1$:

   $\frac{\cos t}{1+\cos t}=\frac{2\cos^2(t/2)-1}{2\cos^2(t/2)}=1-\frac12\sec^2\frac t2.$

   5. Integrate: $x=\displaystyle\int\left(1-\frac12\sec^2\frac t2\right)dt = t-\tan\frac t2+c$.
   6. Substitute back $t=x+y$: $x = (x+y)-\tan\!\left(\dfrac{x+y}{2}\right)+c \Rightarrow 0=y-\tan\!\left(\dfrac{x+y}2\right)+c \Rightarrow y=\tan\!\left(\dfrac{x+y}2\right)-c$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If the length of the sub-tangent at any point P on a curve is proportional to the abscissa of the point P, then the equation of that curve is (C is an arbitrary constant) (A) $y^k+x^k=C$ (B) $x^{1/k}C=y^k$ (C) $(x+y)^k=C$ (D) $y=x^{1/k}C$

   ›Reveal solutionSolution

   "Sub-tangent $\propto$ abscissa" is a separable first-order ODE whose solution is a power curve $y=Cx^{1/k}$.

   Concept and Intuition

   The sub-tangent at a point $P(x,y)$ on a curve is the length, along the x-axis, between the foot of the ordinate at $P$ and the point where the tangent at $P$ meets the x-axis; its formula is $T = \dfrac{y}{dy/dx}$. Saying this length is proportional to $x$ turns the geometric condition into a simple differential equation relating $y$ and $x$.

   Step-by-Step Solution

   1. Sub-tangent: $T = \dfrac{y}{dy/dx}$.
   2. Given condition: $T = kx$ for a proportionality constant $k$, i.e. $\dfrac{y}{dy/dx}=kx$.
   3. Rearranging: $\dfrac{dy}{dx} = \dfrac{y}{kx}$, which separates as $\dfrac{dy}{y} = \dfrac{dx}{kx}$.
   4. Integrate both sides: $\ln y = \dfrac{1}{k}\ln x + \ln C$ (writing the constant of integration as $\ln C$). …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.The general solution of the differential equation $\tan x\tan y\,dx + \cos^2 x\,\csc^2 y\,dy = 0$ is (A) $\tan^2 x + \cot^2 y = C$ (B) $\cot^2 x - \tan^2 y = C$ (C) $\tan^2 x - \cot^2 y = C$ (D) $\cot^2 x + \tan^2 y = C$

   ›Reveal solutionSolution

   This is a variables-separable differential equation; separating and integrating both sides gives the implicit general solution $\tan^2x - \cot^2y = C$.

   Concept and Intuition

   When a first-order differential equation can be written so that all the $x$-terms (with $dx$) are on one side and all the $y$-terms (with $dy$) are on the other, it is separable, and the general solution follows directly by integrating each side independently.

   Step-by-Step Solution

   1. Given: $\tan x\tan y\,dx + \cos^2x\csc^2y\,dy = 0$.
   2. Divide throughout by $\cos^2x\tan y$ (both nonzero on the domain of interest):

   $\frac{\tan x}{\cos^2x}dx + \frac{\csc^2y}{\tan y}dy = 0 \implies \tan x\sec^2x\,dx + \frac{\csc^2y}{\tan y}dy = 0$

   3. Integrate the $x$-term: let $u=\tan x$, $du=\sec^2x\,dx$, so $\int \tan x\sec^2x\,dx = \int u\,du = \frac{u^2}{2} = \frac{\tan^2x}{2}$.
   4. Integrate the $y$-term: $\dfrac{\csc^2y}{\tan y} = \dfrac{1}{\sin^2y}\cdot\dfrac{\cos y}{\sin y} = \dfrac{\cos y}{\sin^3y}$. Let $v=\sin y$, $dv=\cos y\,dy$, so $\int \dfrac{\cos y}{\sin^3y}dy = \int v^{-3}dv = -\dfrac{1}{2v^2} = -\dfrac{1}{2\sin^2y} = -\dfrac{1}{2}(1+\cot^2y)$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.The general solution of $\frac{dy}{dx}=\cos^2(x-y-1)$ is given by $x=$ (A) $C-\cot(x-y-1)$ (B) $C-\tan(x-y+1)$ (C) $y+C\cot(x-y-1)$ (D) $Cy+\tan(x-y-1)$

    ›Reveal solutionSolution

    Substituting $v=x-y-1$ converts the ODE into a directly separable one in $v$ and $x$, giving $x=C-\cot(x-y-1)$.

    Concept and Intuition

    Whenever a first-order ODE has the right-hand side depending only on a linear combination like $x-y-1$ (not $x$ and $y$ separately), substituting $v=x-y-1$ converts it into a separable equation in $v$ alone, since $dv/dx$ becomes a function of $v$ only.

    Step-by-Step Solution

    1. Let $v=x-y-1$, so $\dfrac{dv}{dx}=1-\dfrac{dy}{dx}$.
    2. The given ODE is $\dfrac{dy}{dx}=\cos^2v$, so $\dfrac{dv}{dx}=1-\cos^2v=\sin^2v$.
    3. Separate: $\dfrac{dv}{\sin^2v}=dx\Rightarrow\csc^2v\,dv=dx$.
    4. Integrate: $-\cot v=x+C_1\Rightarrow x=-\cot v-C_1=C-\cot v$ (renaming the constant).
    5. Substitute back $v=x-y-1$: $x=C-\cot(x-y-1)$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.The slope of the tangent at any point $(x,y)$ on the curve, is equal to the product of the coordinates of that point. If the equation of the normal to the curve at the point $(\sqrt2,e)$ is $ax+by=1$, then $\dfrac{b}{a}=$ (A) $\dfrac{1}{\sqrt2\,e}$ (B) $\dfrac{e}{\sqrt2}$ (C) $\sqrt2\,e$ (D) $\dfrac{\sqrt2}{e}$

    ›Reveal solutionSolution

    A separable differential equation gives the curve $y=e^{x^2/2}$; the normal's direction ratios at the given point give $b/a=\sqrt2\,e$.

    Concept and Intuition

    "Slope of tangent equals product of coordinates" is literally $\dfrac{dy}{dx}=xy$ — a separable ODE. Once we know the curve, the tangent slope at any point is just $xy$ evaluated there, and the normal's slope is the negative reciprocal. Writing the normal's equation in the form $ax+by=1$ and comparing coefficients gives $b/a$ directly — no need to fully solve for $a,b$ individually, just their ratio (which is minus the reciprocal of the slope, essentially).

    Step-by-Step Solution

    1. Set up and solve: $\dfrac{dy}{dx}=xy\Rightarrow \displaystyle\int\dfrac{dy}{y}=\int x\,dx\Rightarrow \ln y=\dfrac{x^2}{2}+C$.
    2. Use the point $(\sqrt2,e)$ to find $C$: $\ln e = \dfrac{(\sqrt2)^2}{2}+C\Rightarrow 1=1+C\Rightarrow C=0$. So $\ln y=\dfrac{x^2}{2}$, i.e. $y=e^{x^2/2}$ (consistent: at $x=\sqrt2$, $y=e^{1}=e$. ✓).
    3. Tangent slope at $(\sqrt2,e)$: $\dfrac{dy}{dx}=xy=\sqrt2\cdot e=\sqrt2\,e$.
    4. Normal slope $=-\dfrac{1}{\sqrt2\,e}$. Equation of normal: $y-e=-\dfrac{1}{\sqrt2\,e}(x-\sqrt2)$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.At any point (x,y) on a curve if the length of the subnormal is (x-1) and the curve passes through (1, 2) then the curve is a conic. A vertex of the curve is (A) $(1, 0)$ (B) $(0, 1)$ (C) $(\sqrt{5}, 0)$ (D) $(0, \sqrt{5})$

    ›Reveal solutionSolution

    The subnormal condition $y\,y'=x-1$ integrates to the conic $y^2-(x-1)^2=4$; checking each option against this equation identifies $(0,\sqrt5)$ as the point lying on the curve.

    Concept and Intuition

    For a curve $y=y(x)$, the subnormal at a point is $y\dfrac{dy}{dx}$. Setting this equal to a given expression in $x$ turns the geometric condition into a solvable, separable differential equation whose solution is the actual conic.

    Step-by-Step Solution

    1. Subnormal $=y\dfrac{dy}{dx}=x-1$, so $y\,dy=(x-1)\,dx$.
    2. Integrate both sides: $\dfrac{y^2}{2}=\dfrac{(x-1)^2}{2}+C$, i.e. $y^2-(x-1)^2=2C$.
    3. The curve passes through $(1,2)$: $4-0=2C\Rightarrow 2C=4$. So the curve is $y^2-(x-1)^2=4$, a hyperbola centred at $(1,0)$.
    4. Test each option against this equation:

    • $(1,0)$: $0-0=0\ne4$.
    • $(0,1)$: $1-1=0\ne4$. …

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