Q.Solve the following differential equation: $\frac{dy}{dx} + 2y \tan x = \sin x; \ y=0 \text{ when } x=\frac{\pi}{3}$

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $\cos x \dfrac{dy}{dx} - y\sin x = 6x$, $\left(0 < x < \dfrac{\pi}{2}\right)$ and $y\left(\dfrac{\pi}{3}\right) = 0$ then $y\left(\dfrac{\pi}{6}\right) =$ (A) $\dfrac{-\pi^2}{4\sqrt{3}}$ (B) $\dfrac{-\pi^2}{2}$ (C) $\dfrac{-\pi^2}{2\sqrt{3}}$ (D) $\dfrac{\pi^2}{2\sqrt{3}}$

   ›Reveal solutionSolution

   $\cos x\,y' - y\sin x = \dfrac{d}{dx}(y\cos x) = 6x$, so $y\cos x = 3x^2+C$; using $y(\pi/3)=0$ gives $y(\pi/6)=\dfrac{-\pi^2}{2\sqrt3}$.

   Concept and Intuition

   Many first-order linear ODEs are secretly an exact derivative in disguise. Here $\cos x\,y'-y\sin x$ is precisely the product rule expansion of $\dfrac{d}{dx}(y\cos x)$, so no integrating factor is even needed.

   Step-by-Step Solution

   1. Observe $\dfrac{d}{dx}(y\cos x) = y'\cos x - y\sin x$ — exactly the LHS.
   2. So the equation is $\dfrac{d}{dx}(y\cos x)=6x$.
   3. Integrate: $y\cos x = 3x^2+C$.
   4. Apply $y(\pi/3)=0$: $0\cdot\cos(\pi/3) = 3\left(\dfrac{\pi}{3}\right)^2+C \Rightarrow 0=\dfrac{\pi^2}{3}+C \Rightarrow C=-\dfrac{\pi^2}{3}$.
   5. At $x=\pi/6$: $y\cos(\pi/6) = 3\left(\dfrac{\pi}{6}\right)^2 - \dfrac{\pi^2}{3} = \dfrac{\pi^2}{12}-\dfrac{4\pi^2}{12} = -\dfrac{3\pi^2}{12}=-\dfrac{\pi^2}{4}$. …

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2. 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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3. 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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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-2021-08-20-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 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If y=y(x) is the solution of $\dfrac{dy}{dx}=\dfrac{x-y\cos x}{1+\sin x}$, $y\left(\dfrac{\pi}{2}\right)=\dfrac{\pi^2}{8}$, then $y(\pi)=$ (A) $\dfrac{5\pi^2}{8}$ (B) $\dfrac{7\pi^2}{8}$ (C) $\dfrac{9\pi^2}{8}$ (D) $\dfrac{12\pi^2}{7}$

   ›Reveal solutionSolution

   Recognise the left side of the rearranged ODE as an exact derivative $\dfrac{d}{dx}[y(1+\sin x)]$, integrate directly, use the initial condition to find the constant, then evaluate at $x=\pi$.

   Concept and Intuition

   Many "linear-looking" first-order ODEs are secretly exact derivatives in disguise — recognising the pattern $y'g(x)+y\,g'(x)=\dfrac{d}{dx}[y\,g(x)]$ turns a substitution-heavy linear-ODE problem into direct integration.

   Step-by-Step Solution

   1. Given: $\dfrac{dy}{dx}=\dfrac{x-y\cos x}{1+\sin x}$. Multiply through by $(1+\sin x)$: $(1+\sin x)\dfrac{dy}{dx}+y\cos x=x$.
   2. Note $\dfrac{d}{dx}\left[y(1+\sin x)\right]=y'(1+\sin x)+y\cos x$ — exactly the left side above.
   3. So $\dfrac{d}{dx}\left[y(1+\sin x)\right]=x$. Integrate: $y(1+\sin x)=\dfrac{x^2}{2}+C$.
   4. Apply $y(\pi/2)=\dfrac{\pi^2}{8}$: $1+\sin(\pi/2)=2$, so LHS $=\dfrac{\pi^2}{8}\times 2=\dfrac{\pi^2}{4}$.
   5. RHS at $x=\pi/2$: $\dfrac{(\pi/2)^2}{2}+C=\dfrac{\pi^2}{8}+C$.
   6. Equate: $\dfrac{\pi^2}{4}=\dfrac{\pi^2}{8}+C \Rightarrow C=\dfrac{\pi^2}{8}$.
   7. General solution: $y(1+\sin x)=\dfrac{x^2}{2}+\dfrac{\pi^2}{8}$. …

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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 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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9. 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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10. 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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11. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.If $f'(x) = \tan^2(x) + \cot^2(x)$ and $f\left(\dfrac{\pi}{4}\right) = 0$, then $f(x) =$ ______ (A) $\tan(x) - \cot(x) - x + \dfrac{\pi}{2}$ (B) $\tan(x) - \cot(x) - 2x + \dfrac{\pi}{2}$ (C) $\tan(x) + \cot(x) - 2x + \dfrac{\pi}{2}$ (D) $\sec(x) - \text{cosec}(x) - 2x + \dfrac{\pi}{2}$

    ›Reveal solutionSolution

    Rewrite $\tan^2x+\cot^2x$ using Pythagorean identities, integrate term by term, then fix the constant using $f(\pi/4)=0$.

    Concept and Intuition

    Whenever you see $\tan^2x$ or $\cot^2x$ in something you must integrate, immediately convert using $\tan^2x=\sec^2x-1$ and $\cot^2x=\csc^2x-1$ — these have known antiderivatives ($\tan x$, $-\cot x$), unlike the squared trig functions themselves.

    Step-by-Step Solution

    1. Rewrite: $f'(x) = (\sec^2x-1)+(\csc^2x-1) = \sec^2x+\csc^2x-2$.
    2. Integrate: $f(x) = \int(\sec^2x+\csc^2x-2)\,dx = \tan x - \cot x - 2x + C$.
    3. Apply the condition $f(\pi/4)=0$: $\tan(\pi/4)=1$, $\cot(\pi/4)=1$, so …

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12. 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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