Q.(a) Solve the differential equation $2(y + 3) - xy\dfrac{dy}{dx} = 0$; given that $y(1) = -2$.

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1. CBSE 2026Set 65/1/11 markMCQ
   Q.The integrating factor of differential equation $R\dfrac{dx}{dy} + Px = Q$, where $P$, $Q$, $R$ are functions of $y$, is (A) $e^{\int \frac{P}{Q}\, dy}$ (B) $e^{\int P\, dy}$ (C) $e^{\int \frac{P}{R}\, dy}$ (D) $e^{\int \frac{P}{R}\, dx}$

   ›Reveal solutionSolution

   The key idea is to rewrite the given equation in the standard linear form $\frac{dx}{dy} + P_1 x = Q_1$ by dividing through by $R$, then the integrating factor is $e^{\int P_1\, dy} = e^{\int \frac{P}{R}\, dy}$. The correct option is (C).

   The Integrating Factor (IF) method is a systematic way to solve first-order linear differential equations. The core insight is that we want to multiply the entire equation by a function that turns the left-hand side into the exact derivative of a product — specifically, the derivative of $x$ times some function. This works because if you have an equation of the form $\frac{dx}{dy} + f(y) x = g(y)$, multiplying both sides by $e^{\int f(y)\, dy}$ makes the left side become $\frac{d}{dy}\left( x \cdot e^{\int f(y)\, dy} \right)$, which is then easy to integrate.

   Here, the given equation is $R\frac{dx}{dy} + Px = Q$, where $P$, $Q$, $R$ are functions of $y$ alone. Notice that the coefficient of $\frac{dx}{dy}$ is $R$, not 1. So our first job is to put it into the standard form.

   1. Rewrite in standard linear form. Divide every term by $R$ (assuming $R \neq 0$):

   $\frac{dx}{dy} + \frac{P}{R}\, x = \frac{Q}{R}.$

   Now it matches the pattern $\frac{dx}{dy} + P_1(y)\, x = Q_1(y)$, where $P_1(y) = \frac{P}{R}$ and $Q_1(y) = \frac{Q}{R}$.

   2. Recall the formula for the integrating factor. For a first-order linear ODE in the form $\frac{dx}{dy} + P_1(y)\, x = Q_1(y)$, the integrating factor is

   $\text{IF} = e^{\int P_1(y)\, dy}.$

   This is a standard result — the exponential of the integral of the coefficient of $x$.

   3. Substitute $P_1$ into the formula. Here $P_1(y) = \frac{P}{R}$, so $\text{IF} = e^{\int \frac{P}{R}\, dy}.$ …

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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 65/2/11 markMCQ
   Q.The integrating factor of the differential equation $2x \frac{dy}{dx} - y = 3$ is (A) $\sqrt{x}$ (B) $\frac{1}{\sqrt{x}}$ (C) $e^x$ (D) $e^{-x}$

   ›Reveal solutionSolution

   To find the integrating factor, we first convert the given differential equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$. From this, we identify $P(x) = -\frac{1}{2x}$, and the integrating factor is calculated as $e^{\int P(x) dx}$, which evaluates to $\frac{1}{\sqrt{x}}$.

   The integrating factor method is a powerful technique used to solve first-order linear differential equations. A first-order linear differential equation has the general form:

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

   where $P(x)$ and $Q(x)$ are functions of $x$ (or constants).

   Why the Integrating Factor?

   The core idea is to transform the left-hand side (LHS) of this equation into the derivative of a product. Specifically, we want to make the LHS look like $\frac{d}{dx} (y \cdot \text{some function})$.

   Let's say we multiply the entire equation by a function, $\mu(x)$, which we call the integrating factor:

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

   Now, consider the product rule for differentiation: $\frac{d}{dx} (\mu(x) y) = \mu(x) \frac{dy}{dx} + y \frac{d\mu}{dx}$.

   For our modified LHS to be exactly $\frac{d}{dx} (\mu(x) y)$, we need the term $\mu(x) P(x)y$ to be equal to $y \frac{d\mu}{dx}$.

   This means:

   $\mu(x) P(x) = \frac{d\mu}{dx}$

   This is a separable differential equation for $\mu(x)$. We can rewrite it as:

   $\frac{d\mu}{\mu} = P(x) dx$

   Integrating both sides:

   $\int \frac{d\mu}{\mu} = \int P(x) dx$

   $\log|\mu| = \int P(x) dx$

   Exponentiating both sides (and typically taking the positive value for $\mu(x)$ as a convention, and omitting the constant of integration since any constant factor in $\mu(x)$ would cancel out later):

   $\mu(x) = e^{\int P(x) dx}$

   This $\mu(x)$ is the integrating factor. Once we multiply the original equation by this $\mu(x)$, the LHS becomes $\frac{d}{dx} (\mu(x) y)$, which can then be easily integrated to solve for $y$.

   Let's apply this method to the given problem.

   1. Convert to Standard Form The given differential equation is $2x \frac{dy}{dx} - y = 3$. The standard form for a first-order linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$. To achieve this, we need the coefficient of $\frac{dy}{dx}$ to be 1. We can do this by dividing the entire equation by $2x$:

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

   $\frac{dy}{dx} - \frac{1}{2x} y = \frac{3}{2x}$

   2. Identify $P(x)$ Now, comparing our equation $\frac{dy}{dx} - \frac{1}{2x} y = \frac{3}{2x}$ with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we can identify $P(x)$ and $Q(x)$:

   $P(x) = -\frac{1}{2x}$

   $$ Q(x) = \frac{3}{2x} $$ …
   

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4. 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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5. 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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6. CBSE 2026Set A1 markMCQ
   Q.The integrating factor of the differential equation $(1 + x^2)\frac{dy}{dx} + y = e^{\tan^{-1}x}$ is

   (a) $e^{\tan^{-1}x}$

   (b) $e^{\sin^{-1}x}$

   (c) $\tan^{-1}x$

   (d) $\sin^{-1}x$

   ›Reveal solutionSolution

   Standard form gives $P=\tfrac{1}{1+x^2}$, so I.F. $=e^{\int\frac{dx}{1+x^2}}=e^{\tan^{-1}x}$.

   Divide the equation by $(1+x^2)$ to get the linear form $\dfrac{dy}{dx}+Py=Q$:

   $\dfrac{dy}{dx}+\dfrac{1}{1+x^2}\,y=\dfrac{e^{\tan^{-1}x}}{1+x^2}$, so $P=\dfrac{1}{1+x^2}$.

   …

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7. 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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8. 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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9. 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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10. 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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11. CBSE 2026Set ANNUAL1 markMCQ
    Q.The Integrating factor of differential equation (1 − y²) dy/dx + y·x = ay is

    (a) 1/(y²−1)

    (b) 1/√(y²−1)

    (c) 1/(1−y²)

    (d) 1/√(1−y²)

    ›Reveal solutionSolution

    Written correctly this is a linear equation in $x$ as a function of $y$: $(1-y^2)\dfrac{dx}{dy} + xy = ay$. Its integrating factor works out to $1/\sqrt{1-y^2}$.

    Divide through by $(1-y^2)$ to get the standard linear form $\dfrac{dx}{dy} + P(y)\,x = Q(y)$:

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

    Integrating factor $\text{I.F.} = e^{\int P(y)\,dy} = e^{\int \frac{y}{1-y^2}dy}$.

    Let $u = 1-y^2$, $du = -2y\,dy$: …

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12. CBSE 2026Set ANNUAL1 mark
    Q.Find the integrating factor of the differential equation x dy/dx + y = 10.

    ›Reveal solutionSolution

    Write the equation in the standard linear form $dy/dx+Py=Q$ first, then use $I.F.=e^{\int P\,dx}$.

    $x\dfrac{dy}{dx}+y=10 \;\Rightarrow\; \dfrac{dy}{dx}+\dfrac1x y = \dfrac{10}{x}$

    Here $P=1/x$.

    …

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