Q.Find the equation of a curve passing through the point $(-2, 3)$, given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$.

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If a curve passes through $(1, 2)$ and has the slope of its tangent $1 - \dfrac{1}{x^2}$ at a point $(x, y)$, then the equation of that curve is (A) $y = 3x - \dfrac{1}{x}$ (B) $y = x + \dfrac{1}{x}$ (C) $y = 2x + \dfrac{1}{x} - 1$ (D) $y = x + \dfrac{2}{x} - 1$

   ›Reveal solutionSolution

   Direct integration of the given slope expression, followed by using the point $(1,2)$ to fix the constant, gives $y=x+\dfrac{1}{x}$.

   Concept and Intuition

   When the slope $\dfrac{dy}{dx}$ is given purely as a function of $x$, the curve is found by straightforward integration (no need for separation of variables in a more complex sense) — then a known point on the curve pins down the constant of integration.

   Step-by-Step Solution

   1. $\dfrac{dy}{dx} = 1 - \dfrac{1}{x^2}$.
   2. Integrate both sides: $y = \displaystyle\int\left(1-\dfrac{1}{x^2}\right)dx = x + \dfrac{1}{x} + C$.
   3. Apply the point $(1,2)$: $2 = 1 + 1 + C \Rightarrow C = 0$. …

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2. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.The equation of a curve passing through the point $(0,1)$, given that the slope of the tangent to the curve at any point $(x,y)$ is equal to the sum of the $x$-coordinate, and the product of $x$ and $y$ coordinates at that point, is ________ (A) $y=1-2e^{(x^2/2)}$ (B) $y=-1+2e^{(x^2/2)}$ (C) $y=-1-2e^{(x^2/2)}$ (D) $y=1+2e^{(x^2/2)}$

   ›Reveal solutionSolution

   Translate the word problem into a separable differential equation $\frac{dy}{dx}=x(1+y)$, solve it, and apply the initial point $(0,1)$ to fix the constant — giving $y=-1+2e^{x^2/2}$.

   Concept and Intuition

   "Slope of the tangent" always means $\frac{dy}{dx}$. The sentence describing the slope translates directly into an algebraic expression in $x$ and $y$; recognizing that it factors as $x(1+y)$ makes the equation separable, which is the easiest class of ODE to solve.

   Step-by-Step Solution

   1. Translate: slope $=$ ($x$-coordinate) $+$ (product of $x$ and $y$) $\Rightarrow \dfrac{dy}{dx} = x + xy = x(1+y)$.
   2. Separate variables: $\dfrac{dy}{1+y} = x\,dx$.
   3. Integrate both sides: $\log|1+y| = \dfrac{x^2}{2} + C$.
   4. Apply the point $(0,1)$: $\log|1+1| = 0 + C \Rightarrow C = \log 2$. …

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3. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.If the slope of the tangent at a point $(x, y)$ on a curve is $\dfrac{y - 4}{x - 3}$ and the curve passes through $(4, 3)$, then the point where it cuts the line $y = x$ is (A) $(1, 1)$ (B) $(3, 3)$ (C) $\left(\dfrac{7}{2}, \dfrac{7}{2}\right)$ (D) $\left(-\dfrac{5}{2}, -\dfrac{5}{2}\right)$

   ›Reveal solutionSolution

   The given slope condition is a separable differential equation whose solution is a straight line through $(3,4)$; using the extra point $(4,3)$ pins down that line exactly, and its intersection with $y=x$ is easy to find.

   Concept and Intuition

   "The slope at every point $(x,y)$ equals $\dfrac{y-4}{x-3}$" is really a differential equation, $\dfrac{dy}{dx}=\dfrac{y-4}{x-3}$. This particular form is separable and integrates to a family of straight lines through the fixed point $(3,4)$ — recognizing that shortcut avoids re-deriving calculus each time, but let's still integrate directly to be rigorous.

   Step-by-Step Solution

   1. Separate variables: $\dfrac{dy}{y-4}=\dfrac{dx}{x-3}$.
   2. Integrate both sides: $\log|y-4|=\log|x-3|+C_1$, so $y-4=K(x-3)$ for some constant $K$.
   3. Use the given point $(4,3)$ (the curve passes through it): $3-4=K(4-3)\Rightarrow -1=K$.
   4. So the curve is $y-4=-(x-3)=3-x$, i.e. $y=7-x$. …

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4. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.The equation of the curve passing through the point $\left(0, \dfrac{\pi}{4}\right)$ and satisfying the differential equation $(e^x \tan y)\, dx + \left((1+e^x)\sec^2 y\right)\, dy = 0$ is given by ______ (A) $(1+e^x)\tan y = 2$ (B) $1 + e^x = 2\tan y$ (C) $1 + e^x = 2\sec y$ (D) $(1+e^x)\tan y = k$

   ›Reveal solutionSolution

   A separable first-order ODE; separating and integrating gives a product relation, and the initial point fixes the constant. The answer is $(1+e^x)\tan y=2$.

   Concept and Intuition

   The given equation is exact/separable once we divide through by $\tan y(1+e^x)$: each side then involves only one variable, and each integrates to a logarithm, whose sum is a constant — exponentiating turns the sum of logs into a product equal to a constant.

   Step-by-Step Solution

   1. $(e^x\tan y)\,dx+(1+e^x)\sec^2y\,dy=0$. Divide both sides by $\tan y(1+e^x)$:

   $\dfrac{e^x}{1+e^x}\,dx+\dfrac{\sec^2y}{\tan y}\,dy=0$

   2. Integrate: $\displaystyle\int\dfrac{e^x}{1+e^x}dx=\log(1+e^x)$ and $\displaystyle\int\dfrac{\sec^2y}{\tan y}dy=\log(\tan y)$.
   3. So $\log(1+e^x)+\log(\tan y)=\log C\Rightarrow(1+e^x)\tan y=C$ (renaming the constant). …

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5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The solution of the differential equation $x^2(y+1)\dfrac{dy}{dx} + y^2(x+1)^2 = 0$, when $y(1) = 2$, is (A) $\log|x^2y| = \dfrac{2}{x} + \dfrac{1}{y} + x - 1$ (B) $\log\left|\dfrac{1}{4}x^2y\right| = \dfrac{1}{x} + \dfrac{2}{y} + x - 1$ (C) $\log\left|\dfrac{1}{2}x^2y\right| = \dfrac{1}{x} + \dfrac{1}{y} - x - \dfrac{1}{2}$ (D) $\log\left|\dfrac{1}{3}x^2y\right| = \dfrac{1}{x} + \dfrac{1}{y} - x + \dfrac{1}{2}$

   ›Reveal solutionSolution

   This is a separable ODE; separating and integrating gives an implicit relation, and the initial condition $y(1)=2$ pins the constant to match option (C).

   Concept and Intuition

   $x^2(y+1)\,dy = -y^2(x+1)^2\,dx$ separates cleanly because all the $y$-terms can be gathered on one side (as $\frac{y+1}{y^2}$) and all $x$-terms on the other (as $\frac{(x+1)^2}{x^2}$). Both sides then reduce to standard integrals: $\frac1y+\frac{1}{y^2}$ integrates via power rule and $\log$, and $\frac{(x+1)^2}{x^2}=1+\frac2x+\frac1{x^2}$ likewise.

   Step-by-Step Solution

   1. Rearranging: $x^2(y+1)\,dy=-y^2(x+1)^2\,dx \Rightarrow \dfrac{y+1}{y^2}\,dy=-\dfrac{(x+1)^2}{x^2}\,dx$.
   2. LHS: $\dfrac{y+1}{y^2}=\dfrac1y+\dfrac{1}{y^2}$, so $\displaystyle\int\left(\frac1y+\frac1{y^2}\right)dy=\log|y|-\frac1y+C_1$.
   3. RHS: $\dfrac{(x+1)^2}{x^2}=\dfrac{x^2+2x+1}{x^2}=1+\dfrac2x+\dfrac1{x^2}$, so $\displaystyle-\int\left(1+\frac2x+\frac1{x^2}\right)dx=-x-2\log|x|+\frac1x+C_2$.
   4. Equate: $\log|y|-\dfrac1y=-x-2\log|x|+\dfrac1x+C$.
   5. Rearranged: $\log|y|+2\log|x|=\dfrac1x+\dfrac1y-x+C \Rightarrow \log(x^2y)=\dfrac1x+\dfrac1y-x+C$.
   6. Apply $y(1)=2$: $\log(1\cdot2)=1+\dfrac12-1+C \Rightarrow \ln2=\dfrac12+C \Rightarrow C=\ln2-\dfrac12$.
   7. Substitute back: $\log(x^2y)=\dfrac1x+\dfrac1y-x+\ln2-\dfrac12$, i.e. $\log(x^2y)-\ln2=\dfrac1x+\dfrac1y-x-\dfrac12$.
   8. $\log(x^2y)-\ln2=\log\!\left(\dfrac{x^2y}{2}\right)=\log\left|\dfrac12x^2y\right|$, so finally …

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6. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.The solution of $(1+y^2)\,dx - xy\,dy = 0$, $y(1)=0$ represents a conic. Its eccentricity is (A) $2$ (B) $1/e$ (C) $1$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   Solving the separable ODE with the given initial condition yields the rectangular hyperbola $x^2-y^2=1$, whose eccentricity is $\sqrt2$.

   Concept and Intuition

   Separate the variables in the ODE, integrate, apply the initial condition to fix the constant, and identify the resulting conic. A rectangular hyperbola ($a=b$) always has eccentricity $\sqrt2$ — a fact worth recognising instantly once the conic's form is found.

   Step-by-Step Solution

   1. Given $(1+y^2)dx - xy\,dy = 0 \Rightarrow (1+y^2)dx = xy\,dy \Rightarrow \dfrac{dx}{x} = \dfrac{y\,dy}{1+y^2}$.
   2. Integrate both sides: $\log x = \dfrac12\log(1+y^2) + C$.
   3. Exponentiate: $x = k\sqrt{1+y^2}$ for some constant $k>0$, i.e. $x^2 = k^2(1+y^2)$, or $x^2 - k^2y^2 = k^2$.
   4. Apply $y(1)=0$: at $x=1,y=0$: $1 = k^2(1+0) \Rightarrow k^2=1 \Rightarrow k=1$.
   5. So the curve is $x^2 - y^2 = 1$, a rectangular hyperbola with $a^2=1$, $b^2=1$. …

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7. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.g(x) is an anti derivative of $f(x)=1+2^x\log 2$ and the graph of $y=g(x)$ passes through $\left(-1,\dfrac{1}{2}\right)$. Then the curve meets the Y-axis at (A) (0, 1) (B) (0, 2) (C) (0, -2) (D) (1, 1)

   ›Reveal solutionSolution

   Integrating $f(x)=1+2^x\log2$ gives $g(x)=x+2^x+C$; the given point fixes $C=1$, and evaluating at $x=0$ gives the Y-intercept $(0,2)$.

   Concept and Intuition

   Since $\dfrac{d}{dx}2^x=2^x\log 2$, the antiderivative of $2^x\log2$ is simply $2^x$ — recognizing this avoids unnecessary substitution work.

   Step-by-Step Solution

   1. $g(x)=\int(1+2^x\log2)\,dx=x+2^x+C$.
   2. Use the given point $\left(-1,\tfrac12\right)$: $g(-1)=-1+2^{-1}+C=-1+\tfrac12+C=-\tfrac12+C=\tfrac12\Rightarrow C=1$.
   3. So $g(x)=x+2^x+1$.
   4. Where the curve meets the Y-axis, $x=0$: $g(0)=0+2^0+1=0+1+1=2$. …

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8. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If $\lim_{x\to\infty} y(x) = \dfrac{\pi}{2}$, then the solution of $x^3\sin y\dfrac{dy}{dx}=2$ is $\cos y=$ (A) $\dfrac{3}{x^2}$ (B) $\dfrac{1}{x}$ (C) $\dfrac{1}{x^2}$ (D) $\dfrac{2}{x^3}$

   ›Reveal solutionSolution

   This is a separable ODE; integrate directly and use the given limiting condition at infinity to pin down the constant of integration.

   Concept and Intuition

   $x^3\sin y\,dy/dx=2$ separates cleanly into a function of $y$ times $dy$ equals a function of $x$ times $dx$. The unusual boundary condition (a limit as $x\to\infty$, rather than a value at a finite point) still fixes the constant because both sides of the solution tend to definite limits.

   Step-by-Step Solution

   1. Rewrite: $x^3\sin y\,\dfrac{dy}{dx}=2 \Rightarrow \sin y\,dy=\dfrac{2}{x^3}dx$.
   2. Integrate both sides: $\displaystyle\int \sin y\,dy=\int 2x^{-3}dx \Rightarrow -\cos y = -x^{-2}+C_0$.
   3. Rearranged: $\cos y = x^{-2}-C_0 = \dfrac{1}{x^2}-C$ (renaming the constant). …

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9. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.The particular solution of the differential equation $\dfrac{dx}{dy} = \dfrac{\sin y (1 + y\cot y)}{x\log(x^2 e)}$, $y(1) = 0$ is (A) $y\sin y = x^2 \log x$ (B) $y^2 \sin y = \log x$ (C) $y = \left(\dfrac{e^2}{\sin e}\right)(x-1)$ (D) $y = e^2 \sec x$

   ›Reveal solutionSolution

   Separating variables, both sides integrate to remarkably clean closed forms ($x^2\log x$ and $y\sin y$), and the initial condition kills the constant.

   Concept and Intuition

   $\log(x^2e)=2\log x+1$ simplifies the right-hand denominator nicely, and $\sin y(1+y\cot y)=\sin y+y\cos y$ is exactly the derivative of $y\sin y$ — recognizing these product-rule patterns avoids messy integration.

   Step-by-Step Solution

   1. Given $\dfrac{dx}{dy}=\dfrac{\sin y(1+y\cot y)}{x\log(x^2e)}$, separate variables: $x\log(x^2e)\,dx=\sin y(1+y\cot y)\,dy$.
   2. $\log(x^2e)=\log x^2+\log e=2\log x+1$, so LHS integrand is $x(2\log x+1)$.
   3. $\displaystyle\int x(2\log x+1)\,dx=\int 2x\log x\,dx+\int x\,dx$. By parts, $\int2x\log x\,dx=x^2\log x-\int x\,dx=x^2\log x-\dfrac{x^2}{2}$.
   4. So LHS integral $=x^2\log x-\dfrac{x^2}{2}+\dfrac{x^2}{2}=x^2\log x$ (the $x^2/2$ terms cancel neatly).
   5. RHS: $\sin y(1+y\cot y)=\sin y+y\cos y$ (since $\sin y\cot y=\cos y$). Note $\dfrac{d}{dy}(y\sin y)=\sin y+y\cos y$ exactly.
   6. So $\displaystyle\int(\sin y+y\cos y)\,dy=y\sin y$. …

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10. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.Let $y = Y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y\tan x = 2x + x^2\tan x$, $x \in \left( \frac{-\pi}{2}, \frac{\pi}{2} \right)$, such that $Y(0) = 1$, then ________ (A) $y\left(\frac{\pi}{4}\right) + Y\left(\frac{-\pi}{4}\right) = \frac{\pi^2}{2} + 2$ (B) $y'\left(\frac{\pi}{4}\right) + Y'\left(\frac{-\pi}{4}\right) = -\sqrt{2}$ (C) $y\left(\frac{\pi}{4}\right) - Y\left(\frac{-\pi}{4}\right) = \sqrt{2}$ (D) $y'\left(\frac{\pi}{4}\right) - Y'\left(\frac{-\pi}{4}\right) = \pi - \sqrt{2}$

    ›Reveal solutionSolution

    Solving the first-order linear ODE explicitly gives $Y(x)=x^2+\cos x$; checking each option against this closed form picks out (D). Answer: $\pi-\sqrt2$.

    Concept and Intuition

    This is a standard first-order linear ODE $\frac{dy}{dx}+P(x)y=Q(x)$, solved with an integrating factor. Once we have the explicit closed form for $Y(x)$, we can just plug in and test every option directly instead of guessing.

    Step-by-Step Solution

    1. ODE: $\dfrac{dy}{dx}+y\tan x = 2x+x^2\tan x$. Integrating factor: $\mu=e^{\int\tan x\,dx}=e^{-\log|\cos x|}=\sec x$.
    2. Multiply through by $\sec x$: $\sec x\,\dfrac{dy}{dx}+y\sec x\tan x = 2x\sec x + x^2\sec x\tan x$.
    3. LHS $=\dfrac{d}{dx}(y\sec x)$. Check RHS: $\dfrac{d}{dx}(x^2\sec x)=2x\sec x+x^2\sec x\tan x$ — matches exactly!
    4. So $\dfrac{d}{dx}(y\sec x)=\dfrac{d}{dx}(x^2\sec x)\Rightarrow y\sec x = x^2\sec x + C \Rightarrow y = x^2+C\cos x$.
    5. Apply $Y(0)=1$: $0+C(1)=1\Rightarrow C=1$. So $Y(x)=x^2+\cos x$.
    6. $Y'(x)=2x-\sin x$. Compute $Y'(\pi/4)=\dfrac\pi2-\dfrac{\sqrt2}2$ and $Y'(-\pi/4)=-\dfrac\pi2+\dfrac{\sqrt2}2$. …

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11. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.If $x\log x\dfrac{dy}{dx} + y = \log x^2$ and $y(e) = 0$, then $y(e^2) =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{2}$

    ›Reveal solutionSolution

    This is a first-order linear ODE in $y$; finding the integrating factor $\log x$ and applying $y(e)=0$ gives $y(e^2)=\frac32$.

    Concept and Intuition

    Dividing the given equation by $x\log x$ puts it in the standard linear form $\dfrac{dy}{dx}+P(x)y=Q(x)$, which is always solvable via an integrating factor $\mu=e^{\int P\,dx}$. Recognizing $\log(x^2)=2\log x$ also simplifies the RHS immediately.

    Step-by-Step Solution

    1. Given: $x\log x\,\dfrac{dy}{dx}+y=\log(x^2)=2\log x$.
    2. Divide throughout by $x\log x$: $\dfrac{dy}{dx}+\dfrac{y}{x\log x}=\dfrac{2}{x}$.
    3. This is linear with $P(x)=\dfrac{1}{x\log x}$. Integrating factor:

    $\mu=e^{\int \frac{1}{x\log x}dx}.$

    Let $u=\log x$, $du=dx/x$, so $\int\frac{dx}{x\log x}=\int\frac{du}{u}=\log|u|=\log|\log x|$. Hence $\mu=e^{\log|\log x|}=\log x$ (positive since $x>1$ in this problem).

    4. Multiply the linear ODE by $\mu=\log x$:

    $\log x\cdot\frac{dy}{dx}+\frac{y}{x}=\frac{2\log x}{x}.$

    5. The LHS is exactly $\dfrac{d}{dx}(y\log x)$ (product rule check: $y'\log x+y\cdot\frac1x$ — matches).
    6. Integrate both sides: $y\log x=\displaystyle\int\frac{2\log x}{x}dx$. With $u=\log x$: $\int 2u\,du=u^2=(\log x)^2$. So $y\log x=(\log x)^2+C$. …

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12. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.The particular solution of the differential equation $(1+y^2)dx-xy\,dy=0,\ y(1)=0$ represents (A) a circle (B) a part of parabola (C) a part of ellipse (D) a part of hyperbola

    ›Reveal solutionSolution

    Separating variables and applying the initial condition gives $x^2-y^2=1$, which is a hyperbola.

    Concept and Intuition

    Many first-order separable ODEs, once solved and simplified, reduce to a recognizable conic section. Here, spotting that the equation separates cleanly into $x$ and $y$ parts is the key step; the initial condition then pins down the constant and hence the specific curve (and which branch of it).

    Step-by-Step Solution

    1. Given: $(1+y^2)dx - xy\,dy = 0$.
    2. Rearranging: $(1+y^2)dx = xy\,dy \Rightarrow \dfrac{dx}{x} = \dfrac{y}{1+y^2}dy$.
    3. Integrate both sides: $\log|x| = \dfrac12\log(1+y^2) + C$.
    4. Apply $y(1)=0$: $\log 1 = 0 = \dfrac12\log(1)+C \Rightarrow C=0$.
    5. So $\log x = \dfrac12\log(1+y^2) \Rightarrow 2\log x = \log(1+y^2) \Rightarrow x^2 = 1+y^2$.
    6. Rearranged: $x^2 - y^2 = 1$ — this is the standard equation of a hyperbola. …

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