Q.Find the general solution of the differential equation $\frac{dy}{dx} - y = \cos x$.

Showing the 12 most recent of 15 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The general solution of the differential equation $x^3 \dfrac{dy}{dx} + 3x^2 y = \cos x$ is (A) $y = \dfrac{\sin x}{x^3} + C$ (B) $y = \dfrac{\sin x}{x^3} + Cx$ (C) $y = \dfrac{\sin x}{x^2} + C$ (D) $y = \dfrac{\sin x + C}{x^2}$ (E) $y = \dfrac{\sin x + C}{x^3}$

   ›Reveal solutionSolution

   Recognize the exact derivative $\dfrac{d}{dx}(x^3y)$.

   Since $\dfrac{d}{dx}(x^3y)=x^3\dfrac{dy}{dx}+3x^2y$, the equation $x^3\dfrac{dy}{dx}+3x^2y=\cos x$ becomes

   $\frac{d}{dx}(x^3y)=\cos x.$

   Integrating: $x^3y=\sin x+C$, hence

   $y=\frac{\sin x+C}{x^3}.$

   ✓Final answer

   The correct option is (E).

2. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The integrating factor of the differential equation $2\, dy = (y + \cos x)\, dx$ is (A) $e^{-2x}$ (B) $e^{-x/2}$ (C) $e^{2x}$ (D) $e^{x/2}$ (E) $-e^{x/2}$

   ›Reveal solutionSolution

   Put in standard linear form and read off the integrating factor.

   From $2\,dy=(y+\cos x)\,dx$:

   $2\frac{dy}{dx}=y+\cos x\ \Rightarrow\ \frac{dy}{dx}-\frac12 y=\frac12\cos x.$

   This is linear with $P(x)=-\dfrac12$, so the integrating factor is

   $e^{\int P\,dx}=e^{-x/2}.$

   ✓Final answer

   The correct option is (B).

3. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The integrating factor of the differential equation $\dfrac{dy}{dx} + \dfrac{2y}{20 - x} = 10$ is (A) $\dfrac{1}{(20 - x)^2}$ (B) $\dfrac{1}{20 - x}$ (C) $20 - x$ (D) $\log_e|20 - x|$ (E) $\log_e(20 - x)^2$

   ›Reveal solutionSolution

   The integrating factor is $e^{\int P\,dx}$ with $P=\dfrac{2}{20-x}$, which evaluates to $\dfrac{1}{(20-x)^2}$.

   For $\dfrac{dy}{dx}+P\,y=Q$ with $P=\dfrac{2}{20-x}$,

   $\int P\,dx=\int\dfrac{2}{20-x}\,dx=-2\log|20-x|.$

   Hence IF $=e^{-2\log|20-x|}=|20-x|^{-2}=\dfrac{1}{(20-x)^2}$.

   ✓Final answer

   The correct option is (A).

4. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The solution of the differential equation $(y + x^2)\, dx = x\, dy, x > 0$ is a curve which passes through the point $(1, 0)$. The equation of the curve is (A) $y = x(x + 1)$ (B) $y = x(x - 1)$ (C) $y = x^2(x - 1)$ (D) $y = x^2(x + 1)$ (E) $y = x(x^2 - 1)$

   ›Reveal solutionSolution

   Rewrite as a first-order linear ODE $y'-y/x=x$, use integrating factor $1/x$, then apply the point $(1,0)$.

   From $(y+x^2)dx=x\,dy$: $x\frac{dy}{dx}=y+x^2\Rightarrow \frac{dy}{dx}-\frac{y}{x}=x$.

   Integrating factor $=e^{-\int dx/x}=\frac1x$, so $\frac{d}{dx}\Big(\frac{y}{x}\Big)=1$.

   Thus $\frac{y}{x}=x+C\Rightarrow y=x^2+Cx$.

   At $(1,0)$: $0=1+C\Rightarrow C=-1$, giving $y=x^2-x=x(x-1)$.

   ✓Final answer

   The correct option is (B).

5. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.The integrating factor of the differential equation $\sin x\, dy = \dfrac{1}{2}(\sin 2x + 2y\cos x)\, dx$ is (A) $\sec x$ (B) $\sin x$ (C) $\tan x$ (D) $\cos x$ (E) $\csc x$

   ›Reveal solutionSolution

   The equation reduces to y' - (cot x) y = cos x, giving integrating factor e^{-integral cot x dx} = csc x.

   Concept and Intuition

   Rewrite the equation in standard linear form dy/dx + P(x)y = Q(x); the integrating factor is e^{integral P dx}.

   Step-by-Step Solution

   1. sin x dy = (1/2)(sin2x + 2y cos x) dx; using sin2x = 2 sin x cos x, the RHS = (sin x cos x + y cos x) dx.
   2. So sin x (dy/dx) = sin x cos x + y cos x, i.e. dy/dx = cos x + y cot x.
   3. Standard form: dy/dx - (cot x) y = cos x, so P(x) = -cot x.
   4. IF = e^{integral -cot x dx} = e^{-log|sin x|} = 1/sin x = csc x.

   Common Mistakes

   • Sign error on P(x) that flips csc x to sin x.

   ✓Final answer

   The correct option is (E) — csc x.

   ANSWER: E

6. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The integrating factor of the differential equation $(1+x^2)dy=(1-2xy)dx$ is (A) $x^2+1$ (B) $\log_e(x^2+1)$ (C) $\dfrac{x}{x^2+1}$ (D) $x(x^2+1)$ (E) $\log_e|x|$

   ›Reveal solutionSolution

   Put in standard linear form $\dfrac{dy}{dx}+Py=Q$ and compute $e^{\int P\,dx}$.

   Rewrite $(1+x^{2})dy=(1-2xy)dx$ as

   $\dfrac{dy}{dx}+\dfrac{2x}{1+x^{2}}\,y=\dfrac{1}{1+x^{2}}.$

   Here $P=\dfrac{2x}{1+x^{2}}$, so

   $\text{IF}=e^{\int\frac{2x}{1+x^{2}}dx}=e^{\log_e(1+x^{2})}=1+x^{2}.$

   ✓Final answer

   The correct option is (A).

7. KEAM 2025Set eng-2025-04274 marksMCQ
   Q.The integrating factor of the differential equation $\dfrac{dy}{dx} - 2y = 2x - 3$ is (A) $e^{2x}$ (B) $\dfrac{-1}{2}e^{-2x}$ (C) $\dfrac{1}{2}e^{-2x}$ (D) $\dfrac{1}{2}e^{-2x}$ (E) $e^{-2x}$

   ›Reveal solutionSolution

   [!TLDR]

   With $P(x) = -2$, the integrating factor $e^{\int P\,dx} = e^{-2x}$.

   Concept

   A linear first-order differential equation in the form $\frac{dy}{dx} + P(x)y = Q(x)$ is solved by multiplying through by the integrating factor $\mu(x) = e^{\int P(x)\,dx}$ — the standard method in the NCERT/CBSE differential-equations chapter.

   Solution

   The equation is

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

   Comparing with $\frac{dy}{dx} + P(x)y = Q(x)$, we read off $P(x) = -2$.

   The integrating factor is

   $\mu(x) = e^{\int P\,dx} = e^{\int (-2)\,dx} = e^{-2x}.$

   [!ANSWER]

   (E) $e^{-2x}$

8. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.The integrating factor of $\left(1 + 2e^{-x}\right)\frac{dy}{dx} - 2e^{-x}y = 1 + e^{-x}$ is (A) $2e^{-x}$ (B) $1 + e^{-x}$ (C) $1 - e^{-x}$ (D) $1 - 2e^{-x}$ (E) $1 + 2e^{-x}$

   ›Reveal solutionSolution

   $\int P\,dx=\log(1+2e^{-x})$, so the integrating factor is $1+2e^{-x}$.

   Divide by $(1+2e^{-x})$:

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

   so $P(x)=-\dfrac{2e^{-x}}{1+2e^{-x}}$. With $w=1+2e^{-x}$, $dw=-2e^{-x}dx$:

   $\int P\,dx=\int\frac{dw}{w}=\log|1+2e^{-x}|.$

   Hence IF $=e^{\log(1+2e^{-x})}=1+2e^{-x}$.

   ✓Final answer

   The correct option is (E).

9. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.The integrating factor of the differential equation $x\frac{dy}{dx}+2y=xe^{x}$ is (A) $\log_{e} x$ (B) $\log_{e} 2x$ (C) $x$ (D) $x^{2}$ (E) $2x$

   ›Reveal solutionSolution

   Dividing by $x$ gives $P(x)=\frac{2}{x}$, so the integrating factor is $x^2$.

   Divide $x\frac{dy}{dx}+2y=xe^x$ by $x$: $\frac{dy}{dx}+\frac{2}{x}y=e^x$.

   The integrating factor is $e^{\int\frac{2}{x}\,dx}=e^{2\log x}=x^2$.

   ✓Final answer

   The correct option is (D).

10. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.The general solution of the differential equation $2y\tan x+\dfrac{dy}{dx}=5\sin x$ is (A) $y=5\sec x+C\sec^2 x$ (B) $y=5+C\cos x$ (C) $y=5\cos x+C$ (D) $y=5\cos x+C\cos^2 x$ (E) $y=5\sec^2 x+C\sec x$

    ›Reveal solutionSolution

    Solve the linear ODE with integrating factor $\sec^2 x$.

    Rewrite as $\frac{dy}{dx}+2\tan x\,y=5\sin x$. Integrating factor:

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

    Then

    $\frac{d}{dx}\left(y\sec^2 x\right)=5\sin x\sec^2 x=5\frac{\sin x}{\cos^2 x}.$

    Integrating the right side: $\int 5\frac{\sin x}{\cos^2 x}dx=\frac{5}{\cos x}=5\sec x$. So

    $y\sec^2 x=5\sec x+C\Rightarrow y=5\cos x+C\cos^2 x.$

    ✓Final answer

    The correct option is (D).

11. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.The integrating factor of the differential equation $(3\sin x\cos x)\,dy=(1+3y\sin^2 x)\,dx$, where $0<x<\dfrac{\pi}{2}$, is (A) $\sec x$ (B) $\sin x$ (C) $\tan x$ (D) $\cos x$ (E) $\cot x$

    ›Reveal solutionSolution

    Put in linear form; the coefficient of $y$ gives IF $=\cos x$.

    From $(3\sin x\cos x)dy=(1+3y\sin^2 x)dx$:

    $\frac{dy}{dx}=\frac{1+3y\sin^2 x}{3\sin x\cos x}=\frac{1}{3\sin x\cos x}+\frac{3\sin^2 x}{3\sin x\cos x}y.$

    The $y$-coefficient is $\frac{\sin x}{\cos x}=\tan x$, so

    $\frac{dy}{dx}-\tan x\,y=\frac{1}{3\sin x\cos x}.$

    Integrating factor:

    $\mu=e^{\int -\tan x\,dx}=e^{\log\cos x}=\cos x.$

    ✓Final answer

    The correct option is (D).

12. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.The general solution of $\dfrac{dy}{dx} + y = 5$ is (A) $-\log|5 - y| = x + C$ (B) $-\log|5 - y| = e^x + C$ (C) $(5 - y)^2 = 2x + C$ (D) $y = \log|x + C|$ (E) $\log|x| + C$

    ›Reveal solutionSolution

    The equation is separable: $\frac{dy}{5-y}=dx$, which integrates to $-\log|5-y|=x+C$.

    Write $\frac{dy}{dx}=5-y$, so $\frac{dy}{5-y}=dx$.

    Integrating the left side, $\int\frac{dy}{5-y}=-\log|5-y|$, hence

    $-\log|5-y|=x+C.$

    ✓Final answer

    The correct option is (A).