Q.Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y' = e^x \sin x$.

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.Solve the differential equation: $\dfrac{dy}{dx} = e^{x+y}$ (A) $e^x + e^y = c$ (B) $e^x - e^y = c$ (C) $e^x + e^{-y} = c$ (D) $e^x - e^{-y} = c$

   ›Reveal solutionSolution

   This tests separation of variables on an exponential differential equation; the key subtlety is that separating $e^{x+y}$ gives $e^{-y}dy = e^x dx$, so the $y$-term integrates to $-e^{-y}$, not $-e^{y}$.

   Concept and Intuition

   Whenever $\dfrac{dy}{dx}$ is a product of a function of $x$ alone and a function of $y$ alone, the equation is separable: we move all $y$-terms (including $dy$) to one side and all $x$-terms (including $dx$) to the other, then integrate independently. Here $e^{x+y}=e^x e^y$ splits cleanly this way.

   Step-by-Step Solution

   1. Write $\dfrac{dy}{dx} = e^{x+y} = e^x\,e^y$.
   2. Separate variables: divide both sides by $e^y$ (equivalently multiply by $e^{-y}$):

   $e^{-y}\,dy = e^x\,dx$

   3. Integrate both sides:

   $\int e^{-y}\,dy = \int e^x\,dx$

   $-e^{-y} = e^x + C_1$

   4. Rearrange to collect the constants on one side:

   $e^x + e^{-y} = -C_1 = c$

   where $c$ is an arbitrary constant of integration.

   Common Mistakes …

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2. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.Find the solution of the following differential equation: $\sin^{-1}\left(\dfrac{dy}{dx}\right) = x + y$ (A) $x = \tan(x+y) + \sec(x+y) + c$ (B) $x = \tan(x+y) - \sec(x+y) + c$ (C) $x = \tan(x+y) + \sec^2(x+y) + c$ (D) $x = \tan(x+y) - \sec^2(x+y) + c$

   ›Reveal solutionSolution

   With $t=x+y$, the equation becomes $\dfrac{dt}{dx}=1+\sin t$, which integrates (via the $1-\sin t$ trick) to $x=\tan t-\sec t+c$.

   Concept and Intuition

   $\sin^{-1}(dy/dx)=x+y$ means $\dfrac{dy}{dx}=\sin(x+y)$ — the right side depends only on the combination $x+y$, which is the classic signal to substitute $t=x+y$ and make the equation separable in $t$ and $x$.

   Step-by-Step Solution

   1. Let $t=x+y\Rightarrow \dfrac{dt}{dx}=1+\dfrac{dy}{dx}=1+\sin t$.
   2. Separate: $\dfrac{dt}{1+\sin t}=dx$.
   3. Multiply top and bottom by $(1-\sin t)$: $\dfrac{(1-\sin t)\,dt}{1-\sin^2 t}=\dfrac{(1-\sin t)\,dt}{\cos^2 t}=(\sec^2 t-\sec t\tan t)\,dt$. …

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3. 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 $\int \dfrac{\cos y}{\sin^3y}dy = \int v^{-3}dv = -\dfrac{1}{2v^2} = -\dfrac{1}{2\sin^2y} = -\dfrac{1}{2}(1+\cot^2y)$. …

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4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.The substitution $\dfrac{dy}{dx} = z$, reduces the differential equation $\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} = 0$ to a differential equation whose solution is $z =$ (A) $\log x + C$ (B) $x + C$ (C) $Ae^x$ (D) $x^2 + C$

   ›Reveal solutionSolution

   Substituting $z=dy/dx$ turns the second-order equation into $dz/dx=z$, a separable ODE with solution $z=Ae^x$.

   Concept and Intuition

   This is the standard order-reduction trick: whenever an ODE involves $y$ only through its first and second derivatives (no explicit $y$ term), setting $z=dy/dx$ turns it into a first-order ODE in $z$, which is often much easier to solve (here, purely separable).

   Step-by-Step Solution

   1. Given $\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0$. Let $z=\dfrac{dy}{dx}$, so $\dfrac{d^2y}{dx^2}=\dfrac{dz}{dx}$.
   2. Substituting: $\dfrac{dz}{dx} - z = 0 \Rightarrow \dfrac{dz}{dx}=z$.
   3. Separate variables: $\dfrac{dz}{z}=dx$.
   4. Integrate: $\log z = x+c \Rightarrow z = e^{x+c} = e^c\cdot e^x$. …

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5. 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 …

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6. 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 $v=x-y-1$: $x=C-\cot(x-y-1)$. …

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7. 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: $\dfrac12\cdot\left[-2\cot\left(\dfrac{v}{2}\right)\right]=x+C \Rightarrow -\cot\left(\dfrac{v}{2}\right)=x+C$. …

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8. 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: $v^2\,dv=-2\,dx$. …

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9. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.If $\dfrac{a}{a_1} = \dfrac{b}{b_1}$, then the substitution to be used to solve the differential equation $\dfrac{dy}{dx} = \dfrac{ax+by+c}{a_1x+b_1y+c_1}$ by using separation of variables is (A) $x = x + h,\ y = y + k$ (B) $ax + by = Z$ (C) $y = V(x).x$ (D) $x = at,\ y = bt$

   ›Reveal solutionSolution

   Proportional coefficients ($a/a_1=b/b_1$) mean the substitution $Z=ax+by$ collapses the RHS to a function of $Z$ alone, making the ODE separable.

   Concept and Intuition

   Normally $\dfrac{dy}{dx}=\dfrac{ax+by+c}{a_1x+b_1y+c_1}$ needs the "shift the origin" substitution $x=X+h,\,y=Y+k$ — but that method fails exactly when $a/a_1=b/b_1$ (the two lines are parallel, so no unique intersection point $(h,k)$ exists). In that degenerate case, set $Z=ax+by$; then $a_1x+b_1y=\dfrac{a_1}{a}Z$ (a constant multiple of $Z$), so the whole RHS becomes a function of $Z$ only, and $\dfrac{dZ}{dx}=a+b\dfrac{dy}{dx}$ gives a separable equation in $Z,x$.

   Step-by-Step Solution

   1. Since $a/a_1=b/b_1=k$ (say), $a_1x+b_1y=\dfrac1k(ax+by)$.
   2. Let $Z=ax+by$. Then RHS $=\dfrac{Z+c}{Z/k+c_1}$, purely a function of $Z$. …

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10. 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: $\dfrac{1}{2}\cdot 2\tan\left(\dfrac{u}{2}\right)=y+c \Rightarrow \tan\left(\dfrac{u}{2}\right)=y+c$. …

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11. 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 $\int\left(-\dfrac1y+\dfrac{2}{2y-1}\right)dy=-\ln|y|+\ln|2y-1|=\ln\left|\dfrac{2y-1}{y}\right|$. …

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12. 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 $\int\dfrac{dy}{y^2+y+1}=\dfrac2{\sqrt3}\tan^{-1}\!\left(\dfrac{2y+1}{\sqrt3}\right)$. …

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