Q.Solve the differential equation $\left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\right) \frac{dx}{dy} = 1\ (x \neq 0)$.

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Let the population of a species of birds surviving at a time ' $\boldsymbol{t}$ ' be governed by the differential equation $\frac{d p}{d t}-p=-100$. If $p(0)=50$, then $p\left(-\log _e 2\right)$ is equal to (A) 100 (B) 90 (C) 75 (D) 40

   ›Reveal solutionSolution

   This is a first-order linear ODE solved using an integrating factor. The solution is $p(t)=100-50e^{t}$, and evaluating at $t=-\log 2$ gives $p=75$, so the correct option is (C).

   ---

   We start with the differential equation

   $\frac{dp}{dt} - p = -100$

   and the initial condition $p(0)=50$. The goal is to find $p(-\log 2)$.

   Concept & Intuition

   This is a first-order linear ordinary differential equation of the form $\frac{dp}{dt} + P(t)p = Q(t)$. Here $P(t) = -1$ (constant) and $Q(t) = -100$ (constant). The standard method is to multiply both sides by an integrating factor $\mu(t) = e^{\int P\,dt}$, which turns the left side into a perfect derivative. Then we integrate and apply the initial condition.

   ---

   Step-by-step solution

   1. Find the integrating factor Since $P(t) = -1$,

   $\mu(t) = e^{\int (-1)\,dt} = e^{-t}.$

   2. Multiply the ODE by $\mu(t)$

   $e^{-t}\frac{dp}{dt} - e^{-t}p = -100 e^{-t}.$

   The left side is exactly $\frac{d}{dt}\left(p e^{-t}\right)$ because

   $\frac{d}{dt}\left(p e^{-t}\right) = e^{-t}\frac{dp}{dt} - p e^{-t}.$

   3. Rewrite and integrate

   $\frac{d}{dt}\left(p e^{-t}\right) = -100 e^{-t}.$

   Integrate both sides with respect to $t$:

   $p e^{-t} = \int -100 e^{-t}\, dt = 100 e^{-t} + C,$

   where $C$ is the constant of integration.

   4. Solve for $p(t)$ Multiply through by $e^{t}$:

   $p(t) = 100 + C e^{t}.$

   5. Apply the initial condition $p(0)=50$

   $50 = 100 + C e^{0} \quad\Rightarrow\quad 50 = 100 + C \quad\Rightarrow\quad C = -50.$

   So the particular solution is

   $$p(t) = 100 - 50 e^{t}. …$$

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.

   [!FORMULA] $\text { The solution of } \boldsymbol{d} \boldsymbol{y}=\boldsymbol{\operatorname { c o s }} \boldsymbol{x}(\mathbf{2}-\boldsymbol{y} \boldsymbol{\operatorname { c o s e c }} \boldsymbol{x}) \boldsymbol{d} \boldsymbol{x} \quad \text { where } y=\sqrt{2} \text { when } x=\frac{\boldsymbol{\pi}}{4} \text { is }$

   (A) $y=\sin x+\frac{1}{2} \operatorname{cosec} x$ (B) $y=\tan \left(\frac{x}{2}\right)+\cot \left(\frac{x}{2}\right)$ (C) $y \sin x=\frac{1}{2} \cos 2 x$ (D) $y=\frac{1}{\sqrt{2}} \sec x+\sqrt{2} \cos \left(\frac{x}{2}\right)$

   ›Reveal solutionSolution

   This is a first‑order linear differential equation. Rewriting it in standard form and using an integrating factor yields the solution $y = \sin x + \frac12 \csc x$, which matches option (A).

   We start with

   $dy = \cos x\,(2 - y \csc x)\,dx.$

   Dividing through by $dx$ (treating it as a differential equation in $y$ and $x$) gives

   $\frac{dy}{dx} = \cos x\,(2 - y \csc x) = 2\cos x - y \cos x \csc x.$

   Since $\cos x \csc x = \cot x$, this becomes

   $\frac{dy}{dx} = 2\cos x - y \cot x.$

   Rearranging into standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:

   $\frac{dy}{dx} + (\cot x)\, y = 2\cos x.$

   Why this approach works:

   A first‑order linear ODE is solved by multiplying through by an integrating factor $\mu(x) = e^{\int P\,dx}$. This turns the left side into the derivative of $\mu(x) y$, which we can then integrate directly.

   1. Find the integrating factor

   $\mu(x) = e^{\int \cot x\,dx} = e^{\log|\sin x|} = \sin x.$

   (We take $\sin x > 0$ for the interval containing $x=\pi/4$.)

   2. Multiply the ODE by $\mu(x)$

   $\sin x \frac{dy}{dx} + (\sin x \cot x) y = 2\sin x \cos x.$

   Since $\sin x \cot x = \cos x$, the left side is exactly $\frac{d}{dx}(y \sin x)$. The right side simplifies: $2\sin x \cos x = \sin 2x$.

   So we have

   $\frac{d}{dx}(y \sin x) = \sin 2x.$

   3. Integrate both sides

   $y \sin x = \int \sin 2x \, dx = -\frac12 \cos 2x + C.$

   4. Use the initial condition Given $y = \sqrt{2}$ when $x = \frac{\pi}{4}$:

   $\sqrt{2} \cdot \sin\frac{\pi}{4} = -\frac12 \cos\frac{\pi}{2} + C.$

   $\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, so left side = $\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 1$. …

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3. KCET 2026Set UNKNOWN1 markMCQ
   Q.Integrating factor of the differential equation $(1-x^2)\frac{dy}{dx} - xy = 1$ is (A) $1-x^2$ (B) $\frac{1}{2}\log(1-x^2)$ (C) $\frac{x}{1-x^2}$ (D) $\sqrt{1-x^2}$

   ›Reveal solutionSolution

   Convert $(1-x^2)\frac{dy}{dx} - xy = 1$ into the standard linear form $\frac{dy}{dx} + Py = Q$, then compute the integrating factor $e^{\int P\,dx}$.

   Step 1 — Write in standard linear form

   Divide throughout by $(1-x^2)$:

   $\frac{dy}{dx} - \frac{x}{1-x^2}\,y = \frac{1}{1-x^2}.$

   Comparing with $\frac{dy}{dx} + Py = Q$, we identify

   $P = -\frac{x}{1-x^2}.$

   Step 2 — Integrate $P$

   $\int P\,dx = \int \left(-\frac{x}{1-x^2}\right)dx.$

   Let $u = 1-x^2$, so $du = -2x\,dx$, i.e. $x\,dx = -\frac{1}{2}du$. Then …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.Integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{x^3+y}{x}$ is (A) $\frac{x}{e^x}$ (B) $e^x$ (C) $\frac{e^x}{x}$ (D) $x e^x$

   ›Reveal solutionSolution

   Rearranging to standard linear form gives $P(x)=1-\frac1x$, so the integrating factor is $\dfrac{e^x}{x}$.

   Rewrite the right-hand side:

   $\frac{dy}{dx}+y=\frac{x^3+y}{x}=x^2+\frac{y}{x}.$

   Bring the $y$-terms together:

   $\frac{dy}{dx}+y-\frac{y}{x}=x^2\quad\Longrightarrow\quad \frac{dy}{dx}+\left(1-\frac{1}{x}\right)y=x^2.$ …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.Solve the following differential equation $\cos ^2 x \frac{d y}{d x}+y=\tan x$, given that $y(0)=1$. Hence find $y\left(\frac{\pi}{4}\right)$ (A) 2 (B) $\frac{2}{e}$ (C) e (D) 1

   ›Reveal solutionSolution

   Rewrite as a linear ODE with integrating factor $e^{\tan x}$: $y=\tan x-1+2e^{-\tan x}$, so $y(\tfrac{\pi}{4})=\tfrac{2}{e}$.

   Divide the equation $\cos^2 x\,\dfrac{dy}{dx}+y=\tan x$ by $\cos^2 x$:

   $\frac{dy}{dx} + \sec^2 x\;y = \tan x\,\sec^2 x$

   This is linear with integrating factor

   $\mu = e^{\int \sec^2 x\,dx} = e^{\tan x}$

   Then

   $\frac{d}{dx}\!\left(y\,e^{\tan x}\right) = \tan x\,\sec^2 x\;e^{\tan x}$

   Put $t=\tan x$, $dt=\sec^2 x\,dx$:

   $y\,e^{\tan x} = \int t\,e^{t}\,dt = e^{t}(t-1) + C = e^{\tan x}(\tan x - 1) + C$

   Hence

   $y = \tan x - 1 + C\,e^{-\tan x}$ …

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.The solution of $(x+\log y) d y+y d x=0$ when $y(0)=1$ is (A) $y(x-1+\log y)+1=0$ (B) $x y+y \log y+1=0$ (C) $x y=y \log y-y-1$ (D) $y(x+1+\log y)-1=0$

   ›Reveal solutionSolution

   Treat as a linear ODE in $x$ with respect to $y$; the integrating factor is $y$, giving $y(x-1+\log y)+1=0$ — option (A).

   Write the equation with $x$ as the dependent variable of $y$:

   $y\,dx + (x+\log y)\,dy = 0 \;\Rightarrow\; \frac{dx}{dy} + \frac{1}{y}\,x = -\frac{\log y}{y}.$

   This is linear in $x$. Integrating factor:

   $\mu = e^{\int \frac{1}{y}\,dy} = e^{\log y} = y.$

   Multiply through:

   $\frac{d}{dy}(x\,y) = -\log y.$

   Integrate, using $\int \log y\,dy = y\log y - y$:

   $x\,y = -(y\log y - y) + C = -y\log y + y + C.$ …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\text { Integrating factor of the differential equation } \frac{d y}{d x}+y=\frac{x^3+y}{x} \text { is }$

   (A) $\frac{x}{e^x}$ (B) $e^x$ (C) $\frac{e^x}{x}$ (D) $x e^x$

   ›Reveal solutionSolution

   Collecting the $y$-terms puts the equation in standard linear form with $P(x)=1-\frac{1}{x}$, giving integrating factor $\frac{e^x}{x}$ — option (C).

   Reduce to standard linear form

   The standard first-order linear form is $\frac{dy}{dx}+P(x)\,y=Q(x)$, with integrating factor $\text{I.F.}=e^{\int P\,dx}$.

   Starting from $\frac{dy}{dx}+y=\frac{x^3+y}{x}$, multiply through by $x$:

   $x\frac{dy}{dx}+xy=x^3+y.$

   Move the $y$ on the right to the left:

   $x\frac{dy}{dx}+(x-1)y=x^3.$

   Divide by $x$:

   $\frac{dy}{dx}+\frac{x-1}{x}\,y=x^2,\qquad P(x)=\frac{x-1}{x}=1-\frac{1}{x}.$

   Compute the integrating factor …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.

   [!FORMULA] $\text { The general solution of the differential equation }(1+\tan y)(d x-d y)+2 x d y=0 \text { is }$

   (A) $y(\sin x+\cos x)=\sin x+c e^x$ (B) $y(\sin x+\cos x)=\sin x+c e^{-x}$ (C) $x(\sin y+\cos y)=\sin y+c e^y$ (D) $x(\sin y+\cos y)=\sin y+c e^{-y}$

   ›Reveal solutionSolution

   This is a first-order linear ODE in disguise. By rewriting the equation in terms of $x$ as a function of $y$, we can apply the integrating factor method. The solution matches option (D).

   We start with

   $(1+\tan y)(dx - dy) + 2x\,dy = 0.$

   The presence of both $dx$ and $dy$ suggests we can treat $x$ as a function of $y$ (or vice versa). Since the options involve $x$ multiplied by trigonometric functions of $y$, it’s natural to solve for $x$ in terms of $y$.

   Why this approach works:

   If we rearrange to isolate $\frac{dx}{dy}$, the equation becomes linear in $x$:

   $\frac{dx}{dy} + P(y)\,x = Q(y).$

   Then we use the standard integrating factor $\mu(y) = e^{\int P\,dy}$.

   ---

   1. Rewrite the equation Expand the given expression:

   $(1+\tan y)\,dx - (1+\tan y)\,dy + 2x\,dy = 0.$

   Group the $dx$ and $dy$ terms:

   $(1+\tan y)\,dx + \bigl[2x - (1+\tan y)\bigr]\,dy = 0.$

   2. Divide by $dy$ to get $\frac{dx}{dy}$ Assuming $dy \neq 0$,

   $(1+\tan y)\,\frac{dx}{dy} + 2x - (1+\tan y) = 0.$

   So

   $(1+\tan y)\,\frac{dx}{dy} + 2x = 1+\tan y.$

   3. Make it linear in $x$ Divide through by $1+\tan y$ (valid where $\tan y \neq -1$):

   $\frac{dx}{dy} + \frac{2}{1+\tan y}\,x = 1.$

   This is a first-order linear ODE:

   $\frac{dx}{dy} + P(y)\,x = Q(y), \quad P(y)=\frac{2}{1+\tan y},\; Q(y)=1.$

   4. Find the integrating factor

   $\mu(y) = e^{\int P\,dy} = e^{\int \frac{2}{1+\tan y}\,dy}.$

   We need $\int \frac{dy}{1+\tan y}$. Use the identity $\tan y = \frac{\sin y}{\cos y}$:

   $\frac{1}{1+\tan y} = \frac{1}{1+\frac{\sin y}{\cos y}} = \frac{\cos y}{\cos y + \sin y}.$

   So

   $\int \frac{2}{1+\tan y}\,dy = 2\int \frac{\cos y}{\cos y + \sin y}\,dy.$

   5. Evaluate the integral A standard trick: Write

   $\cos y = \frac{1}{2}\bigl[(\cos y + \sin y) + (\cos y - \sin y)\bigr].$

   Then

   $\frac{\cos y}{\cos y + \sin y} = \frac{1}{2} + \frac{1}{2}\cdot\frac{\cos y - \sin y}{\cos y + \sin y}.$

   The second term’s integral is easy because the numerator is the derivative of the denominator (up to sign):

   $\frac{d}{dy}(\cos y + \sin y) = -\sin y + \cos y = \cos y - \sin y.$

   Thus

   $\int \frac{\cos y - \sin y}{\cos y + \sin y}\,dy = \log|\cos y + \sin y| + C.$

   Therefore

   $2\int \frac{\cos y}{\cos y + \sin y}\,dy = 2\left(\frac{y}{2} + \frac{1}{2}\log|\cos y + \sin y|\right) = y + \log|\cos y + \sin y|.$

   So the integrating factor is

   $\mu(y) = e^{y + \log|\cos y + \sin y|} = e^y \cdot |\cos y + \sin y|.$

   We can drop the absolute value (absorb sign into constant later) and take

   $\mu(y) = e^y (\cos y + \sin y).$

   6. Apply the integrating factor Multiply the ODE by $\mu(y)$:

   $e^y(\cos y + \sin y)\,\frac{dx}{dy} + e^y(\cos y + \sin y)\cdot\frac{2}{1+\tan y}\,x = e^y(\cos y + \sin y).$

   But note:

   $\frac{2}{1+\tan y} = \frac{2\cos y}{\cos y + \sin y},$

   so the left side becomes

   $e^y(\cos y + \sin y)\,\frac{dx}{dy} + 2e^y\cos y\,x.$

   Observe that

   $\frac{d}{dy}\bigl[e^y(\cos y + \sin y)\,x\bigr] = e^y(\cos y + \sin y)\,\frac{dx}{dy} + x\cdot\frac{d}{dy}\bigl[e^y(\cos y + \sin y)\bigr].$

   Compute the derivative:

   $\frac{d}{dy}\bigl[e^y(\cos y + \sin y)\bigr] = e^y(\cos y + \sin y) + e^y(-\sin y + \cos y) = e^y(2\cos y).$ …

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.The particular solution of the differential equation $\cos x \frac{d y}{d x}+y=\sin x$ at $y(0)=1$ (A) $y(\sec x+\tan x)=\sec x+\tan x-x+1$ (B) $y(\sec x+\tan x)=\sec x+\tan x-x$ (C) $y(\sec x+\tan x)=\sec x+\tan x+x$ (D) $y(\sec x+\tan x)=\sec x+\tan x-x+2$

   ›Reveal solutionSolution

   Solve the linear ODE with integrating factor $\sec x+\tan x$; applying $y(0)=1$ gives $C=0$, so $y(\sec x+\tan x)=\sec x+\tan x-x$ (option B).

   Rewrite the equation in standard linear form by dividing through by $\cos x$:

   $\frac{dy}{dx}+y\sec x=\tan x$

   The integrating factor is

   $\text{IF}=e^{\int \sec x\,dx}=e^{\log|\sec x+\tan x|}=\sec x+\tan x$

   Multiplying through:

   $\frac{d}{dx}\big[y(\sec x+\tan x)\big]=\tan x(\sec x+\tan x)=\sec x\tan x+\tan^2 x$

   Using $\tan^2 x=\sec^2 x-1$:

   $\frac{d}{dx}\big[y(\sec x+\tan x)\big]=\sec x\tan x+\sec^2 x-1$

   Integrating both sides:

   $y(\sec x+\tan x)=\sec x+\tan x-x+C$ …

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10. COMEDK 2023Set 2023-E1 markMCQ
    Q.The general solution of the differential equation $\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y$ (A) $x=\tan ^{-1} y-1+c e^{\tan ^{-1} y}$ (B) $x=\tan ^{-1} y-1+c e^{-\tan ^{-1} y}$ (C) $x=\tan ^{-1} y+c e^{\tan ^{-1} y}$ (D) $x=c \tan ^{-1} y+e^{-\tan ^{-1} y}$

    ›Reveal solutionSolution

    This is linear in $x$ with integrating factor $e^{\tan^{-1}y}$; solving gives $x=\tan^{-1}y-1+ce^{-\tan^{-1}y}$.

    Rewrite $(1+y^2)\,dx=(\tan^{-1}y-x)\,dy$ as

    $\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{\tan^{-1}y}{1+y^2}.$

    Integrating factor: $\mu=e^{\int \frac{dy}{1+y^2}}=e^{\tan^{-1}y}$. Then

    $x\,e^{\tan^{-1}y}=\int \frac{\tan^{-1}y}{1+y^2}\,e^{\tan^{-1}y}\,dy.$

    Put $t=\tan^{-1}y$, $dt=\frac{dy}{1+y^2}$: …

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11. COMEDK 2023Set 2023-E1 markMCQ
    Q.The solution of the differential equation $\frac{d y}{d x}+y \cos x=\frac{1}{2} \sin 2 x$ (A) $y e^{\sin x}=e^{\sin x}(\sin x+1)+c$ (B) $y e^{\sin x}=e^{\sin x}(\sin x-1)+c$ (C) $y e^{\sin 2 x}=e^{\sin 2 x}(\sin x-1)+c$ (D) $y e^{\cos x}=e^{\sin x}(\cos x-1)+c$

    ›Reveal solutionSolution

    Back-substitute t = sin x: y e^(sin x) = e^(sin x) (sin x - 1) + c.

    Concept: linear first-order ODE, dy/dx + P(x) y = Q(x), solved by the integrating factor e^(integral P dx).

    Here P = cos x, Q = (1/2) sin 2x = sin x cos x.

    IF = e^(integral cos x dx) = e^(sin x).

    Solution: y * e^(sin x) = integral [ sin x cos x * e^(sin x) ] dx.

    Put t = sin x, dt = cos x dx: …

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12. COMEDK 2021Set 20211 markMCQ
    Q.The solution of the differential equation $(1 + {y^2}) + (x - {e^{{{\tan }^{ - 1}}y}}){{dy} \over {dx}} = 0$ is (A) $2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + C$ (B) $x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + C$ (C) $x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + C$ (D) $(x - 2) = C{e^{ - {{\tan }^{ - 1}}y}}$

    ›Reveal solutionSolution

    Hence x e^(arctan y) = (1/2) e^(2 arctan y) + C', i.e. 2x e^(arctan y) = e^(2 arctan y) + C.

    Concept: treat x as the dependent variable and y as the independent variable; it becomes a linear first-order ODE in x.

    (1 + y^2) + (x - e^(arctan y)) dy/dx = 0

    => (1 + y^2) dx/dy + x - e^(arctan y) = 0

    => dx/dy + x/(1 + y^2) = e^(arctan y)/(1 + y^2).

    Integrating factor: IF = exp( integral dy/(1 + y^2) ) = e^(arctan y).

    Multiply through:

    d/dy [ x e^(arctan y) ] = e^(2 arctan y)/(1 + y^2).

    Integrate the RHS with t = arctan y, dt = dy/(1 + y^2): …

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