Q.Solve the following differential equation: $\frac{dy}{dx} + y = 1 (y \neq 1)$

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

   $p(t) = 100 - 50 e^{t}.$

   6. Evaluate at $t = -\log 2$

   $p(-\log 2) = 100 - 50 e^{-\log 2}.$

   Recall $e^{-\log 2} = \frac{1}{2}$. Therefore,

   $p(-\log 2) = 100 - 50 \cdot \frac{1}{2} = 100 - 25 = 75.$

   Tip

   A quick check: as $t \to -\infty$, $e^{t} \to 0$, so $p(t) \to 100$. That makes sense — the population approaches the equilibrium $p=100$. At $t=0$ we start at 50, so at a negative time (going backward) the population should be between 50 and 100, and 75 fits perfectly.

   Watch out

   A common mistake is to mis-handle the sign in the integrating factor: $\int (-1)\,dt = -t$, not $+t$. Also, be careful with $e^{-\log 2} = 1/2$, not $-1/2$.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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 = $\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 1$.

   $\cos\frac{\pi}{2} = 0$, so $1 = 0 + C$ ⇒ $C = 1$.

   5. Write the explicit solution

   $y \sin x = -\frac12 \cos 2x + 1.$

   Using the identity $\cos 2x = 1 - 2\sin^2 x$, we get

   $y \sin x = -\frac12(1 - 2\sin^2 x) + 1 = -\frac12 + \sin^2 x + 1 = \sin^2 x + \frac12.$

   Hence

   $y = \sin x + \frac12 \csc x.$

   Tip

   The identity $\cos 2x = 1 - 2\sin^2 x$ is the cleanest way to simplify; alternatively, one could use $\cos 2x = 2\cos^2 x - 1$ but that leads to a less tidy form.

   Watch out

   A common mistake is forgetting the constant of integration or misapplying the initial condition. Always check that the final expression satisfies both the ODE and the given point.

   Comparing with the options, this matches (A).

   ✓Final answer

   The correct option is (A).

   ANSWER: A

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

   $\int \left(-\frac{x}{1-x^2}\right)dx = \int \left(-\frac{1}{u}\right)\left(-\frac{1}{2}\right)du = \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(1-x^2).$

   Step 3 — Form the integrating factor

   $I.F. = e^{\int P\,dx} = e^{\frac{1}{2}\ln(1-x^2)} = (1-x^2)^{1/2} = \sqrt{1-x^2}.$

   ✓Final answer

   The correct option is (D) — $\sqrt{1-x^2}$.

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:

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

   This is linear with $P(x)=1-\dfrac1x$. The integrating factor is

   $\text{I.F.}=e^{\int\left(1-\frac1x\right)dx}=e^{\,x-\log x}=\frac{e^{x}}{x}.$

   ✓Final answer

   Integrating factor $=\dfrac{e^{x}}{x}$, which is option (C).

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

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

   Apply $y(0)=1$ (with $\tan 0 = 0$):

   $1 = (0-1) + C \implies C = 2$

   $y = \tan x - 1 + 2\,e^{-\tan x}$

   At $x=\dfrac{\pi}{4}$, $\tan\dfrac{\pi}{4}=1$:

   $y\!\left(\frac{\pi}{4}\right) = 1 - 1 + 2e^{-1} = \frac{2}{e}$

   ✓Final answer

   $y\!\left(\dfrac{\pi}{4}\right) = \dfrac{2}{e}$ — option (B).

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

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

   Apply $y(0)=1$ (i.e. $x=0,\;y=1$):

   $0 = -1\cdot 0 + 1 + C \;\Rightarrow\; C = -1.$

   Hence

   $xy = -y\log y + y - 1 \;\Rightarrow\; xy + y\log y - y + 1 = 0 \;\Rightarrow\; y\,(x-1+\log y) + 1 = 0.$

   ✓Final answer

   $y(x-1+\log y)+1=0$ — option (A).

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

   $\text{I.F.}=e^{\int\left(1-\frac{1}{x}\right)dx}=e^{x-\log x}=\frac{e^x}{e^{\log x}}=\frac{e^x}{x}.$

   The most common slip is to read the coefficient of $y$ as $1$ and get $e^x$; that ignores the $y/x$ hidden on the right-hand side.

   ✓Final answer

   Integrating factor $=\dfrac{e^x}{x}$. Correct option: (C).

   ANSWER: C

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:

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

   So indeed the left side is exactly $\frac{d}{dy}\bigl[e^y(\cos y + \sin y)\,x\bigr]$.

   Thus the equation becomes

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

   7. Integrate both sides

   $e^y(\cos y + \sin y)\,x = \int e^y(\cos y + \sin y)\,dy + C.$

   The integral on the right is a standard form:

   $\int e^y(\cos y + \sin y)\,dy = e^y \sin y + \text{constant},$

   because $\frac{d}{dy}(e^y \sin y) = e^y \sin y + e^y \cos y = e^y(\sin y + \cos y)$.

   So we have

   $e^y(\cos y + \sin y)\,x = e^y \sin y + C.$

   8. Solve for $x$ Divide through by $e^y(\cos y + \sin y)$ (nonzero in general):

   $x = \frac{\sin y}{\cos y + \sin y} + \frac{C}{e^y(\cos y + \sin y)}.$

   Multiply numerator and denominator by something? The options are written as $x(\sin y + \cos y) = \sin y + c e^{-y}$. Multiply both sides of our result by $(\cos y + \sin y)$:

   $x(\cos y + \sin y) = \sin y + C e^{-y}.$

   This matches option (D) exactly, with $c = C$.

   Watch out

   A common mistake is to try solving for $y$ in terms of $x$ first, which leads to a much messier nonlinear equation. Recognizing the linear structure in $x$ as a function of $y$ is key.

   Tip

   The integral $\int \frac{dy}{1+\tan y}$ can also be done by substituting $u = y + \pi/4$, using $\tan(y) = \tan(u - \pi/4)$, but the “split numerator” method shown is faster.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

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:

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

   Apply the initial condition $y(0)=1$. At $x=0$, $\sec 0=1$ and $\tan 0=0$, so the left side is $1\cdot(1+0)=1$ and the right side is $1+0-0+C=1+C$. Hence $C=0$.

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

   ✓Final answer

   The particular solution is $y(\sec x+\tan x)=\sec x+\tan x-x$ — option (B).

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

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

    Hence $x\,e^{t}=e^{t}(t-1)+c$, i.e.

    $x=\tan^{-1}y-1+c\,e^{-\tan^{-1}y}.$

    ✓Final answer

    The correct option is (B) — $x=\tan ^{-1} y-1+c e^{-\tan ^{-1} y}$

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:

    integral t e^t dt = t e^t - e^t + c = e^t (t - 1) + c.

    Back-substitute t = sin x:

    y e^(sin x) = e^(sin x) (sin x - 1) + c.

    ✓Final answer

    The correct option is (B) — $y e^{\sin x}=e^{\sin x}(\sin x-1)+c$

    ANSWER: B

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):

    integral e^(2t) dt = e^(2t)/2 = e^(2 arctan y)/2.

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

    2x e^(arctan y) = e^(2 arctan y) + C.

    ✓Final answer

    The correct option is (A) — $2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + C$

    ANSWER: A