Q.Solve the following differential equation: $\frac{dy}{dx} + (\sec x) y = \tan x \quad \left(0 \le x < \frac{\pi}{2}\right)$

Showing the 12 most recent of 36 on this concept.

1. CBSE 2023Set 65/2/11 markMCQ
   Q.The integrating factor of the differential equation $(1 - y^2)\frac{dx}{dy} + yx = ay$, $(-1 < y < 1)$ is:

   (a) $\frac{1}{y^2 - 1}$

   (b) $\frac{1}{\sqrt{y^2 - 1}}$

   (c) $\frac{1}{1 - y^2}$

   (d) $\frac{1}{\sqrt{1 - y^2}}$

   ›Reveal solutionSolution

   The key idea is to rewrite the equation in the standard linear form $\frac{dx}{dy} + P(y)x = Q(y)$ and then compute the integrating factor as $\mu(y) = e^{\int P(y)\,dy}$. For this problem, the integrating factor is $\frac{1}{\sqrt{1 - y^2}}$, which corresponds to option (d).

   The Integrating Factor (IF) method is the standard tool for solving first-order linear differential equations. The equation is linear in $x$ as a function of $y$, so we aim to write it in the form $\frac{dx}{dy} + P(y)x = Q(y)$. Once in that form, the IF is $e^{\int P\,dy}$ — it works because multiplying through by the IF turns the left side into the exact derivative of $(x \cdot \text{IF})$, making integration straightforward.

   Let’s go step by step.

   1. Rewrite the given equation in standard linear form. The equation is $(1 - y^2)\frac{dx}{dy} + yx = ay$. Divide through by $(1 - y^2)$ (valid since $-1 < y < 1$, so $1 - y^2 > 0$):

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

   Now it matches $\frac{dx}{dy} + P(y)x = Q(y)$ with $P(y) = \frac{y}{1 - y^2}$.

   2. Find the integrating factor. The integrating factor is $\mu(y) = e^{\int P(y)\,dy} = e^{\int \frac{y}{1 - y^2}\,dy}$. Compute the integral: let $u = 1 - y^2$, so $du = -2y\,dy$, hence $y\,dy = -\frac{1}{2}du$. Then

   $\int \frac{y}{1 - y^2}\,dy = \int \frac{1}{u}\left(-\frac{1}{2}\right)du = -\frac{1}{2}\log|u| + C = -\frac{1}{2}\log(1 - y^2) + C,$

   since $1 - y^2 > 0$ in the given interval, so absolute value is unnecessary.

   Therefore,

   $\mu(y) = e^{-\frac{1}{2}\log(1 - y^2)} = e^{\log\left((1 - y^2)^{-1/2}\right)} = \frac{1}{\sqrt{1 - y^2}}.$ …

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2. CBSE 2025Set 65/1/11 markMCQ
   Q.The integrating factor of differential equation $(x + 2y^3) \frac{dy}{dx} = 2y$ is (A) $e^{\frac{y^2}{2}}$ (B) $\frac{1}{\sqrt{y}}$ (C) $\frac{1}{y^2}$ (D) $e^{-\frac{1}{y^2}}$

   ›Reveal solutionSolution

   The given equation is not linear in $y$, but it is linear in $x$ when rewritten as $\frac{dx}{dy} - \frac{x}{2y} = y^2$. The integrating factor is $e^{\int -\frac{1}{2y}\,dy} = \frac{1}{\sqrt{y}}$, so the correct option is (B).

   We often learn the integrating factor method for first-order linear ODEs of the form $\frac{dy}{dx} + P(x)y = Q(x)$. But here, the equation is $(x + 2y^3) \frac{dy}{dx} = 2y$. If we try to write it as $\frac{dy}{dx} + P(x)y = Q(x)$, we run into trouble because the coefficient of $\frac{dy}{dx}$ involves both $x$ and $y$, and the right-hand side is $2y$ — not a function of $x$ alone. That path leads nowhere.

   The trick is to swap the roles of $x$ and $y$. Notice that the equation contains $x$ and $y$ in a way that suggests $x$ might be the dependent variable. If we rewrite it as $\frac{dx}{dy}$, we get a linear equation in $x$ — and then the integrating factor method works cleanly.

   Let’s do it step by step.

   1. Rewrite the equation in terms of $\frac{dx}{dy}$. Start with

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

   Divide both sides by $\frac{dy}{dx}$ (which is fine as long as $y$ is not constant):

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

   Now isolate $\frac{dx}{dy}$:

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

   2. Bring it to the standard linear form. A first-order linear ODE in $x$ (as a function of $y$) looks like

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

   From $\frac{dx}{dy} = \frac{x}{2y} + y^2$, subtract $\frac{x}{2y}$ from both sides:

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

   So here $P(y) = -\frac{1}{2y}$ and $Q(y) = y^2$.

   3. Find the integrating factor. The integrating factor for $\frac{dx}{dy} + P(y)x = Q(y)$ is

   $\mu(y) = e^{\int P(y)\,dy}.$

   With $P(y) = -\frac{1}{2y}$, we have

   $\int -\frac{1}{2y}\,dy = -\frac{1}{2} \log|y| = \log\left(y^{-1/2}\right).$ …

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3. CBSE 2024Set 65/1/11 markMCQ
   Q.The integrating factor of the differential equation $(1 - x^2) \frac{dy}{dx} + xy = ax$, $-1 < x < 1$ is : (A) $\frac{1}{x^2 - 1}$ (B) $\frac{1}{\sqrt{x^2 - 1}}$ (C) $\frac{1}{1 - x^2}$ (D) $\frac{1}{\sqrt{1 - x^2}}$

   ›Reveal solutionSolution

   The differential equation is linear in $y$, so we rewrite it in standard form $\frac{dy}{dx} + P(x)y = Q(x)$ and compute the integrating factor $\mu(x) = e^{\int P(x)\,dx}$. The correct integrating factor is $\frac{1}{\sqrt{1 - x^2}}$, which corresponds to option (D).

   We are given:

   $(1 - x^2) \frac{dy}{dx} + xy = ax, \quad -1 < x < 1$

   This is a first-order linear differential equation in $y$. The standard form is:

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

   The integrating factor (I.F.) is then:

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

   Why does this work?

   The idea is to multiply the entire equation by $\mu(x)$ so that the left-hand side becomes the derivative of $\mu(x) y$. This turns the problem into a direct integration. The key is finding $P(x)$ correctly — and that means isolating $\frac{dy}{dx}$ with coefficient 1.

   ---

   1. Rewrite in standard form Divide every term by $(1 - x^2)$:

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

   So here:

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

   2. Find $\int P(x)\,dx$ We need:

   $\int \frac{x}{1 - x^2}\, dx$

   Let $u = 1 - x^2$, then $du = -2x\,dx$, so $x\,dx = -\frac{1}{2} du$.

   The integral becomes:

   $\int \frac{x}{1 - x^2}\, dx = \int \frac{1}{u} \left(-\frac{1}{2}\right) du = -\frac{1}{2} \log |u| + C = -\frac{1}{2} \log |1 - x^2| + C$

   Since $-1 < x < 1$, we have $1 - x^2 > 0$, so absolute values are unnecessary:

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

   3. Compute the integrating factor $$$$

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4. 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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5. 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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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 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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8. 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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9. CBSE 2026Set ANNUAL1 mark
   Q.Write the answer in one word/sentence: Write the general solution of a differential equation of the type $\frac{dx}{dy} + P_1 x = Q_1$.

   ›Reveal solutionSolution

   Linear DE in $x$; multiply by integrating factor $e^{\int P_1\,dy}$.

   For $\dfrac{dx}{dy}+P_1x=Q_1$, the integrating factor is $\text{IF}=e^{\int P_1\,dy}$, and the solution is

   …

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10. CBSE 2025Set ANNUAL1 mark
    Q.The integrating factor of differential equation $\dfrac{dy}{dx}+y=x$ is _____.

    ›Reveal solutionSolution

    For a linear equation $\dfrac{dy}{dx}+Py=Q$, the integrating factor is $e^{\int P\,dx}$.

    Here $\dfrac{dy}{dx}+y=x$ is linear with $P=1$, $Q=x$.

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.Integrating factor of differential equation $\dfrac{dy}{dx} - \dfrac{y}{x} = x^4 - 3x$ is

    (a) $x$

    (b) $\log x$

    (c) $\dfrac{1}{x}$

    (d) $-x$

    ›Reveal solutionSolution

    The integrating factor of a linear DE depends only on the coefficient of y, not on the right-hand-side function of x — so it is unaffected by the exact power in the RHS term.

    The equation $\dfrac{dy}{dx} - \dfrac{y}{x} = x^4 - 3x$ is a first-order linear differential equation of the standard form $\dfrac{dy}{dx} + P(x)y = Q(x)$, with:

    $P(x) = -\dfrac{1}{x}$, and $Q(x) = x^4-3x$ (the right-hand side, whatever its exact power).

    The integrating factor is:

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

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.The integrating factor of ydx − (x + 2y²)dy = 0 is :

    (a) 1/y²

    (b) −1/y

    (c) 1/y

    (d) y

    ›Reveal solutionSolution

    Rewritten as a linear equation in $x$ (treating $y$ as the independent variable), the integrating factor is $e^{\int -\frac1y\,dy}=\dfrac1y$.

    Given: $y\,dx-(x+2y^2)\,dy=0$, i.e. $y\,dx = (x+2y^2)\,dy$, so

    $\frac{dx}{dy} = \frac{x}{y}+2y \quad\Longrightarrow\quad \frac{dx}{dy} - \frac1y\,x = 2y.$

    …

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