Q.Find the equation of a curve passing through the point $(0, 2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Showing the 12 most recent of 26 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $y = f(x)$ is a solution of $\dfrac{dy}{dx} = (y - Kx)^2 + K$ when $x = 0$ and $y = 1$, then $f(2) =$ (A) $2K + 1$ (B) $2K - 1$ (C) $2K + 5$ (D) $2K - 5$

   ›Reveal solutionSolution

   A shift substitution $v=y-Kx$ turns the equation into the separable $dv/dx=v^2$; solving and evaluating at $x=2$ gives $f(2)=2K-1$.

   Concept and Intuition

   The right side $(y-Kx)^2+K$ is not separable in $y$ directly, but the combination $y-Kx$ appears squared — this is the signal to substitute $v=y-Kx$, which typically removes the extra additive constant $K$ from the derivative and leaves a clean separable equation in $v$.

   Step-by-Step Solution

   1. Let $v=y-Kx$. Then $\dfrac{dv}{dx}=\dfrac{dy}{dx}-K$.
   2. Given $\dfrac{dy}{dx}=(y-Kx)^2+K=v^2+K$, so $\dfrac{dv}{dx}=v^2+K-K=v^2$.
   3. Separate variables: $\dfrac{dv}{v^2}=dx\Rightarrow -\dfrac{1}{v}=x+c$.
   4. Initial condition: at $x=0$, $y=1\Rightarrow v=y-K\cdot0=1$. So $-1=0+c\Rightarrow c=-1$.
   5. Thus $-\dfrac1v=x-1\Rightarrow v=\dfrac{1}{1-x}$. …

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10. 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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11. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.If the solution of $\dfrac{dy}{dx} = \dfrac{y^3 \cos\sqrt{x}}{\sqrt{x}\, e^{1/y^2}}$, $y(0) = 1$ is $\dfrac{1}{y^2} = \log_e(f(x))$, then $f(x) =$ (A) $4 + 4\sin\sqrt{x}$ (B) $e\sin\sqrt{x}$ (C) $1 - 4\sin\sqrt{x}$ (D) $e - 4\sin\sqrt{x}$

    ›Reveal solutionSolution

    This is a separable ODE; separating variables and using the initial condition $y(0)=1$ gives $f(x)=e-4\sin\sqrt x$.

    Concept and Intuition

    The equation separates cleanly into a function of $y$ times $dy$ equal to a function of $x$ times $dx$. Both sides then need simple substitutions ($u=1/y^2$ and $v=\sqrt x$) to become directly integrable exponential/trig forms.

    Step-by-Step Solution

    1. $\dfrac{dy}{dx} = \dfrac{y^3\cos\sqrt x}{\sqrt x\,e^{1/y^2}} \Rightarrow e^{1/y^2}\,y^{-3}\,dy = \dfrac{\cos\sqrt x}{\sqrt x}\,dx$.
    2. LHS: let $u=1/y^2 \Rightarrow du = -2y^{-3}dy \Rightarrow y^{-3}dy = -\tfrac12 du$. So $\displaystyle\int e^u\left(-\tfrac12\right)du = -\tfrac12 e^u + C_1 = -\tfrac12 e^{1/y^2}+C_1$.
    3. RHS: let $v=\sqrt x \Rightarrow dv=\dfrac{dx}{2\sqrt x}\Rightarrow \dfrac{dx}{\sqrt x}=2\,dv$. So $\displaystyle\int\cos v\cdot2\,dv = 2\sin v + C_2 = 2\sin\sqrt x + C_2$.
    4. Combine: $-\dfrac12e^{1/y^2} = 2\sin\sqrt x + C$.
    5. Apply $y(0)=1 \Rightarrow 1/y^2=1$ at $x=0$: $-\dfrac12 e^1 = 2\sin0+C \Rightarrow C=-\dfrac{e}{2}$. …

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12. 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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