Q.Find the equation of the curve passing through the point $\left(0, \frac{\pi}{4}\right)$ whose differential equation is $\sin x \cos y\, dx + \cos x \sin y\, dy = 0$.

Showing the 12 most recent of 44 on this concept.

1. CBSE 20241 markMCQ
   Q.The solution of the differential equation $\frac{dy}{dx} = 1 - x + y - xy$ is : (A) $\log |1 + y| = x - \frac{x^2}{2} + c$ (B) $\log |1 + y| = -x + \frac{x^2}{2} + c$ (C) $e^y = x - \frac{x^2}{2} + c$ (D) $e^{(1+y)} = -x + \frac{x^2}{2} + c$

   ›Reveal solutionSolution

   The equation $\frac{dy}{dx} = 1 - x + y - xy$ can be factored and solved by separation of variables. The correct solution is $\log |1 + y| = x - \frac{x^2}{2} + c$, which corresponds to option (A).

   The key to solving this is noticing that the right-hand side can be grouped into factors, each depending on only one variable. That’s the signal to use Separation of Variables — a method where we rewrite the equation so that all $y$ terms are on one side and all $x$ terms on the other, then integrate both sides.

   Let’s see why this works here.

   1. Factor the right-hand side The given equation is:

   $\frac{dy}{dx} = 1 - x + y - xy$

   Group the terms cleverly:

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

   Factor out $(1 - x)$:

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

   Now the derivative equals a product of a function of $x$ and a function of $y$. That’s the perfect setup for separation.

   2. Separate the variables Multiply both sides by $dx$ and divide by $(1 + y)$ (assuming $1 + y \neq 0$ for now):

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

   The variables are now isolated — $y$ on the left, $x$ on the right.

   3. Integrate both sides

   $\int \frac{dy}{1 + y} = \int (1 - x)\,dx$

   The left integral is a standard logarithmic form:

   $\log |1 + y| = x - \frac{x^2}{2} + C$

   where $C$ is the constant of integration. …

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2. CBSE 2026Set 65/2/11 markMCQ
   Q.The general solution of the differential equation $\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}$ is (A) $\log \sqrt{y} = \log \sqrt{x} + C$ (B) $\sqrt{y} + \sqrt{x} = C$ (C) $\sqrt{y} - \sqrt{x} = C$ (D) $\log \sqrt{y} + \log \sqrt{x} = C$

   ›Reveal solutionSolution

   This is a separable first-order ODE. Separate variables, integrate, and simplify to get $\sqrt{y} - \sqrt{x} = C$, which is option (C).

   The key idea: whenever you see $\frac{dy}{dx}$ expressed as a ratio of functions of $y$ and $x$ alone, you can "separate" them — move all $y$ terms to one side and all $x$ terms to the other — then integrate each side independently. That's exactly what we have here: $\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}$.

   Notice that $\sqrt{y}$ depends only on $y$, and $\sqrt{x}$ only on $x$. So the equation is already in separable form — we just need to rearrange it properly.

   1. Separate the variables. Multiply both sides by $dx$ and divide by $\sqrt{y}$:

   $\frac{dy}{\sqrt{y}} = \frac{dx}{\sqrt{x}}$

   This is valid as long as $x > 0$ and $y > 0$ (so the square roots are defined and non-zero).

   2. Integrate both sides. Each side is a standard power integral:

   $\int y^{-1/2}\, dy = \int x^{-1/2}\, dx$

   $\frac{y^{1/2}}{1/2} = \frac{x^{1/2}}{1/2} + C_1$

   which simplifies to:

   $2\sqrt{y} = 2\sqrt{x} + C_1$

   3. Simplify the constant. Divide through by 2:

   $\sqrt{y} = \sqrt{x} + \frac{C_1}{2}$

   Let $C = \frac{C_1}{2}$ (just renaming the arbitrary constant). Then:

   $\sqrt{y} - \sqrt{x} = C$

   Watch out

   A common mistake is to forget the constant of integration or to combine the two integration constants incorrectly. When you integrate both sides, you get a constant on each side — but they can be merged into a single constant. Also, don't lose the factor of 2 from the power rule: $\int u^{-1/2}\, du = 2u^{1/2}$, not $u^{1/2}$. …

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3. CBSE 2026Set 65/3/11 markMCQ
   Q.Questions number 19 and 20 are Assertion-Reason based questions. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A): A particular solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is $e^x + e^{-y} = -2$. Reason (R): The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is $e^x + e^{-y} = C$.

   ›Reveal solutionSolution

   Separate variables in $\frac{dy}{dx} = e^{x+y}$ to find the general solution $e^x + e^{-y} = C$; the particular solution $e^x + e^{-y} = -2$ is impossible because the left side is always positive while the right is negative.

   The differential equation $\frac{dy}{dx} = e^{x+y}$ is separable. The key insight is to rewrite the exponential sum in the exponent as a product: $e^{x+y} = e^x \cdot e^y$. This lets us collect all $x$-terms with $dx$ and all $y$-terms with $dy$.

   Solving the differential equation

   1. Separate the variables.

   $\frac{dy}{dx} = e^x \cdot e^y$

   Rearranging:

   $\frac{dy}{e^y} = e^x \, dx$

   or equivalently,

   $e^{-y} \, dy = e^x \, dx$

   2. Integrate both sides.

   $\int e^{-y} \, dy = \int e^x \, dx$

   The left side gives $-e^{-y}$ and the right gives $e^x$:

   $-e^{-y} = e^x + C_1$

   where $C_1$ is an arbitrary constant.

   3. Rearrange to standard form. Multiply through by $-1$:

   $e^{-y} = -e^x - C_1$

   or equivalently,

   $e^x + e^{-y} = -C_1$

   Renaming $-C_1$ as $C$ (still an arbitrary constant):

   $e^x + e^{-y} = C$

   This is the general solution, so Reason (R) is true.

   Checking the particular solution

   Now examine the proposed particular solution $e^x + e^{-y} = -2$.

   For this to be valid, we need $C = -2$ in the general solution. But notice:

   • $e^x > 0$ for all real $x$ …

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4. CBSE 2026Set CX1 markMCQ
   Q.The solution of $\dfrac{dy}{dx}=e^{x+y}$ is:

   (a) $e^{-y}=e^{x}+c$

   (b) $e^{x}+e^{-y}=c$

   (c) $e^{-x}-e^{-y}=c$

   (d) $e^{-x}+e^{-y}=c$

   ›Reveal solutionSolution

   The equation is variable-separable; separating and integrating gives $e^{x}+e^{-y}=c$ — option (b).

   Why separate? Since $e^{x+y}=e^{x}\,e^{y}$, the right side factors into an $x$-part and a $y$-part, so the variables separate cleanly.

   $\dfrac{dy}{dx}=e^{x}\,e^{y}\;\Rightarrow\;e^{-y}\,dy=e^{x}\,dx$

   …

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5. CBSE 2026Set A1 markMCQ
   Q.The solution of differential equation $\frac{dy}{dx} = e^{x+y}$ is

   (a) $e^x + e^{-y} = k$

   (b) $e^x + e^y = k$

   (c) $e^{-x} + e^y = k$

   (d) $e^{-x} + e^{-y} = k$

   ›Reveal solutionSolution

   Separate variables in $\tfrac{dy}{dx}=e^{x+y}$: $\int e^{-y}dy=\int e^x dx\Rightarrow e^x+e^{-y}=k$.

   Write $\dfrac{dy}{dx}=e^{x+y}=e^x\cdot e^y$ and separate:

   $e^{-y}\,dy=e^x\,dx$.

   …

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6. CBSE 2026Set A1 markMCQ
   Q.The solution of differential equation $x\frac{dy}{dx} = \cot y$ is

   (a) $x\cos y = k$

   (b) $x\tan y = k$

   (c) $x\sec y = k$

   (d) $x\sin y = k$

   ›Reveal solutionSolution

   Separate variables: $\int\tan y\,dy=\int\tfrac{dx}{x}\Rightarrow \log|\sec y|=\log|x|+c\Rightarrow x\cos y=k$.

   From $x\dfrac{dy}{dx}=\cot y$, separate:

   $\dfrac{dy}{\cot y}=\dfrac{dx}{x}\Rightarrow \tan y\,dy=\dfrac{dx}{x}$.

   …

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7. CBSE 2026Set ANNUAL1 mark
   Q.The general solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ is __________.

   ›Reveal solutionSolution

   Separate variables and integrate both sides using the standard integral of $1/(1+t^2)$.

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

   …

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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} = e^{x+y}$ is:

   (a) $e^x + e^{-y} = c$

   (b) $e^x + e^y = c$

   (c) $e^{-x} + e^y = c$

   (d) $e^{-x} + e^{-y} = c$

   ›Reveal solutionSolution

   Separate variables using $e^{x+y}=e^x\cdot e^y$ and integrate both sides.

   $\dfrac{dy}{dx}=e^{x+y}=e^x\cdot e^y$

   Separating variables: $e^{-y}\,dy=e^x\,dx$

   …

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9. CBSE 2026Set ANNUAL1 mark
   Q.Find the general solution of the differential equation \frac{dy}{dx} = (1 + x^2)(1 + y^2).

   ›Reveal solutionSolution

   $\tan^{-1}y=x+\dfrac{x^3}{3}+c$.

   Concept. A separable differential equation $\dfrac{dy}{dx}=g(x)h(y)$ is solved by collecting all $y$-terms on one side and all $x$-terms on the other, then integrating both sides.

   Steps.

   • $\dfrac{dy}{dx}=(1+x^2)(1+y^2)$.
   • Separate: $\dfrac{dy}{1+y^2}=(1+x^2)\,dx$. …

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10. CBSE 2026Set ANNUAL1 markMCQ
    Q.The solution of the differential equation $(x^{2}+1)\dfrac{dy}{dx}=1,\ y(1)=\dfrac{\pi}{2}$ is

    (a) $y=\tan^{-1}x+\dfrac{\pi}{3}$

    (b) $y=\tan^{-1}x$

    (c) $y=\tan^{-1}x+\dfrac{\pi}{6}$

    (d) $y=\tan^{-1}x+\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    Integrating gives $y=\tan^{-1}x+C$; the condition fixes $C=\tfrac{\pi}{4}$.

    Step 1: $(x^{2}+1)\dfrac{dy}{dx}=1\Rightarrow dy=\dfrac{dx}{x^{2}+1}$.

    Step 2: Integrating, $y=\tan^{-1}x+C$.

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.The general solution to the differential equation $\dfrac{y\,dx - x\,dy}{y} = 0$ is:

    (a) $y = cx$

    (b) $x = cy^2$

    (c) $xy = c$

    (d) $y = cx^2$

    ›Reveal solutionSolution

    Separate variables and integrate.

    $\dfrac{y\,dx-x\,dy}{y}=0 \implies y\,dx = x\,dy \implies \dfrac{dx}{x}=\dfrac{dy}{y}$

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.The general solution of the differential equation $\dfrac{dy}{dx} = e^{x+y}$ is

    (a) $e^x + e^y = C$

    (b) $e^{-x} + e^y = C$

    (c) $e^x + e^{-y} = C$

    (d) $e^{-x} + e^{-y} = C$

    ›Reveal solutionSolution

    Separate variables ($e^{-y}dy = e^x dx$) and integrate both sides.

    $\dfrac{dy}{dx} = e^{x+y} = e^x\cdot e^y$

    Separating variables:

    $e^{-y}\,dy = e^x\,dx$

    Integrating both sides: …

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