Q.The solution of $\frac{dy}{dx}+y=e^{-x}$, $y(0)=0$ is:
(A) $y=e^{-x}(x-1)$
(B) $y=xe^x$
(C) $y=xe^{-x}+1$
(D) $y=xe^{-x}$

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Given that $\dfrac{dy}{dx}=ye^{x}$ is such that when $x=0, y=e$. The value of $y$ $(y>0)$ when $x=1$ will be (A) $\dfrac{1}{e}$ (B) $e$ (C) $e^{e}$ (D) $\log e$

   ›Reveal solutionSolution

   This is a separable first-order ODE; separating and integrating gives $\log y=e^x+c$, and the initial condition pins $c=0$, so $y(1)=e^e$.

   Concept and Intuition

   When a differential equation can be written as $\dfrac{dy}{y}=g(x)\,dx$, both sides can be integrated independently, giving $\log y = G(x)+c$ where $G$ is the antiderivative of $g$. The given initial condition then fixes the constant, after which we simply plug in the target $x$-value.

   Step-by-Step Solution

   1. Given $\dfrac{dy}{dx}=ye^x$, separate variables: $\dfrac{dy}{y}=e^x\,dx$.
   2. Integrate both sides: $\log y = e^x + c$ (using $y>0$, so no absolute value needed).
   3. Apply the initial condition $x=0,\ y=e$: $\log e = e^0+c\Rightarrow 1=1+c\Rightarrow c=0$. …

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2. 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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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If the solution of $\dfrac{dy}{dx} = \log(x + 1)$ when $y(0) = 3$ is $y = (x + 1)\log(x + 1) + f(x)$, then $f(x) =$ (A) $3 - x$ (B) $x - 3$ (C) $1 - x$ (D) $x - 1$

   ›Reveal solutionSolution

   This tests integrating $\log(x+1)$ by substitution and then pinning the constant with the initial condition. The answer is $f(x)=3-x$.

   Concept and Intuition

   The differential equation $\dfrac{dy}{dx}=\log(x+1)$ is already separated — $y$ is just the antiderivative of the right-hand side. The initial condition $y(0)=3$ fixes the single constant of integration, and comparing the resulting expression to the given form of the solution lets us read off $f(x)$ without guessing.

   Step-by-Step Solution

   1. Integrate: $y=\displaystyle\int \log(x+1)\,dx + C$.
   2. Substitute $u=x+1$, $du=dx$: $\displaystyle\int \log u\,du = u\log u - u + C$ (standard result, by parts with $dv=du$).
   3. Back-substitute: $y=(x+1)\log(x+1)-(x+1)+C$.
   4. Compare with the given form $y=(x+1)\log(x+1)+f(x)$: so $f(x) = -(x+1)+C = C-1-x$. …

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4. 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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5. 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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6. 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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7. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.If the general solution of the differential equation $\cos^2 x\,\dfrac{dy}{dx} + y = \tan x$ is $y = \tan x - 1 + Ce^{-\tan x}$ satisfies $y(\pi/4) = 1$, then $C =$ (A) $e$ (B) $1$ (C) $-1$ (D) $1/e$

   ›Reveal solutionSolution

   This tests applying an initial condition to a given general solution of a linear first-order DE to find the arbitrary constant.

   Concept and Intuition

   Once a general solution $y=y(x,C)$ is known, an initial condition (initial value) pins down $C$ by substituting the given point directly — no need to re-derive the DE.

   Step-by-Step Solution

   1. General solution: $y=\tan x-1+Ce^{-\tan x}$.
   2. At $x=\pi/4$: $\tan(\pi/4)=1$.
   3. So $y(\pi/4)=1-1+Ce^{-1}=Ce^{-1}$. …

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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 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If f(x) is a function such that $f'(x)=\sqrt{f^2(x)-1}$ and $f(0)=1$, then $f(1)=$ (A) $\frac{e^{-2}+1}{2e}$ (B) $\frac{e^2+1}{2e}$ (C) $\frac{e^2-1}{2e}$ (D) $\frac{e^{-2}-1}{2e}$

   ›Reveal solutionSolution

   This is a separable differential equation whose natural substitution is the hyperbolic identity $\cosh^2u-\sinh^2u=1$; the answer is $f(1)=\cosh 1=\dfrac{e^2+1}{2e}$.

   Concept and Intuition

   Whenever you see $f'(x)=\sqrt{f^2(x)-1}$, the structure $\sqrt{f^2-1}$ screams hyperbolic substitution, because $\cosh^2u-1=\sinh^2u$. Setting $f=\cosh u$ turns the messy square root into the clean function $\sinh u$, and the chain rule collapses the whole ODE to $u'=1$ — a straight line in disguise.

   Step-by-Step Solution

   1. Separate variables: $\dfrac{df}{\sqrt{f^2-1}}=dx$.
   2. Let $f=\cosh u$, so $df=\sinh u\,du$ and $\sqrt{f^2-1}=\sinh u$ (taking $u\ge 0$ since $f\ge 1$ near $x=0$).
   3. The equation becomes $\dfrac{\sinh u\,du}{\sinh u}=dx\Rightarrow du=dx\Rightarrow u=x+C$. …

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10. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.If the solution of $\frac{dy}{dx} - y\log_e 0.5 = 0$, $y(0) = 1$, and $y(x) \to k$, as $x \to \infty$ then $k =$ (A) $\infty$ (B) $-1$ (C) $1$ (D) $0$

    ›Reveal solutionSolution

    A linear first-order ODE $dy/dx = ky$ with $k=\log_e 0.5<0$ describes exponential decay, so $y\to0$ as $x\to\infty$; $k=0$.

    Concept and Intuition

    $\dfrac{dy}{dx}=cy$ always solves to $y=y_0e^{cx}$. The sign of $c$ decides growth ($c>0$) or decay ($c<0$) as $x\to\infty$. Here $c=\log_e(0.5)=-\log_e2 <0$, so the solution decays to zero.

    Step-by-Step Solution

    1. Rewrite: $\dfrac{dy}{dx}=y\log_e(0.5)$, i.e. $\dfrac{dy}{y}=\log_e(0.5)\,dx$.
    2. Integrate: $\log_e y = x\log_e(0.5)+C_1$, so $y=Ce^{x\log_e(0.5)}=C(0.5)^x$.
    3. Apply $y(0)=1$: $1=C(0.5)^0=C$, so $C=1$. Hence $y(x)=(0.5)^x$. …

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11. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
    Q.Let the differential equation for which $y = Ae^x + Be^{-x}$ is the general solution is of degree $m$ and order $n$. If the above differential equation satisfies the condition $y(0) = m$ and $y'(0) = n$, then its solution is $y =$ (A) $\dfrac{e^x + e^{-x}}{2}$ (B) $\dfrac{e^x - e^{-x}}{2}$ (C) $\dfrac{3e^x - e^{-x}}{2}$ (D) $\dfrac{e^x - 3e^{-x}}{2}$

    ›Reveal solutionSolution

    The general solution $y=Ae^x+Be^{-x}$ solves $y''-y=0$ (order 2, degree 1); using $y(0)=1,\ y'(0)=2$ pins down $y=\tfrac{3e^x-e^{-x}}{2}$.

    Concept and Intuition

    Before using the initial conditions we must correctly identify $m$ (degree) and $n$ (order) of the differential equation whose general solution is $y=Ae^x+Be^{-x}$. Since this has 2 arbitrary constants, the DE must be second order; and since it can be written with the highest derivative appearing to the first power ($y''=y$), the degree is 1.

    Step-by-Step Solution

    1. From $y=Ae^x+Be^{-x}$: $y'=Ae^x-Be^{-x}$, $y''=Ae^x+Be^{-x}=y$. So the DE is $y''-y=0$: order $n=2$, degree $m=1$.
    2. Given conditions: $y(0)=m=1$, $y'(0)=n=2$.
    3. $y(0)=A+B=1$.
    4. $y'(0)=A-B=2$.
    5. Adding: $2A=3\Rightarrow A=\tfrac32$. Subtracting: $2B=1-2=-1\Rightarrow B=-\tfrac12$. …

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12. 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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