Q.Solve the following differential equation: $\frac{dy}{dx} + 3y = e^{-2x}$

Showing the 12 most recent of 36 on this concept.

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

   $\text{IF} = e^{\int \frac{P}{R}\, dy}.$

   Watch out

   A common mistake is to forget to divide by $R$ first and then incorrectly use $P$ directly. The integrating factor depends on the coefficient of $x$ after the derivative term has coefficient 1. Also, note that the integration is with respect to $y$ (the independent variable), not $x$ — option (D) integrates with respect to $x$, which is wrong because $P$ and $R$ are functions of $y$.

   Tip

   If you ever forget the formula, just remember: you want to find a function $\mu(y)$ such that $\mu \frac{dx}{dy} + \mu P_1 x = \frac{d}{dy}(\mu x)$. Expanding the right side gives $\mu \frac{dx}{dy} + \frac{d\mu}{dy} x$, so comparing, we need $\frac{d\mu}{dy} = \mu P_1$, which gives $\mu = e^{\int P_1\, dy}$.

   Thus, the integrating factor is $e^{\int \frac{P}{R}\, dy}$.

   ✓Final answer

   The correct option is (C).

2. 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 $Q(x)$:

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

   $Q(x) = \frac{3}{2x}$

   > [!WARNING]
   > Be careful with the sign! $P(x)$ includes the sign of the $y$ term. Here, it's $ - \frac{1}{2x} $, not just $ \frac{1}{2x} $.
   

   3. Calculate the Integrating Factor

   The formula for the integrating factor (IF) is:

   > [!FORMULA]

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

   Substitute $P(x) = -\frac{1}{2x}$ into the formula:

   $IF = e^{\int -\frac{1}{2x} dx}$

   Now, we need to evaluate the integral in the exponent:
   

   $\int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx$

   $= -\frac{1}{2} \log|x|$

   For typical exam problems involving integrating factors, we usually assume $x > 0$ (or the domain where the solution is valid) so we can write $ \log x $.
   

   $= -\frac{1}{2} \log x$

   Using the logarithm property $a \log b = \log(b^a)$:
   

   $= \log(x^{-1/2})$

   Now, substitute this back into the integrating factor formula:
   

   $IF = e^{\log(x^{-1/2})}$

   Using the property $e^{\log A} = A$:
   

   $IF = x^{-1/2}$

   $IF = \frac{1}{\sqrt{x}}$

   Comparing this result with the given options:

   (A) $\sqrt{x}$

   (B) $\frac{1}{\sqrt{x}}$

   (C) $e^x$

   (D) $e^{-x}$

   Our calculated integrating factor matches option (B).

   ✓Final answer

   The integrating factor of the differential equation is $\boxed{\frac{1}{\sqrt{x}}}$.

3. 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 $P=\dfrac{1}{1+x^2}$.

   The integrating factor is $e^{\int P\,dx}=e^{\int\frac{dx}{1+x^2}}=e^{\tan^{-1}x}$.

   ✓Final answer

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

4. 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 $u = 1-y^2$, $du = -2y\,dy$:

   $\int \frac{y}{1-y^2}dy = -\frac12\int \frac{du}{u} = -\frac12\ln|1-y^2|.$

   So $\text{I.F.} = e^{-\frac12\ln|1-y^2|} = (1-y^2)^{-1/2} = \dfrac{1}{\sqrt{1-y^2}}.$

   ✓Final answer

   $\dfrac{1}{\sqrt{1-y^2}}$. (Option d)

5. 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 $P=1/x$.

   $I.F. = e^{\int P\,dx} = e^{\int \frac1x\,dx} = e^{\ln x} = x$

   ✓Final answer

   Integrating factor $= x$.

6. 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

   $x\cdot\text{IF}=\displaystyle\int Q_1\cdot\text{IF}\,dy+C.$

   ✓Final answer

   $x\,e^{\int P_1\,dy}=\displaystyle\int Q_1\,e^{\int P_1\,dy}\,dy+C$.

7. 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

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

   Therefore

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

   Tip

   Notice that we never needed to solve for $x$ — the question only asks for the integrating factor. Once you see the equation is linear in $x$, the integrating factor comes directly from $P(y)$.

   4. Check the options. The integrating factor we found is $\frac{1}{\sqrt{y}}$, which matches option (B).

   Watch out

   A common mistake is to try to force the equation into the form $\frac{dy}{dx} + P(x)y = Q(x)$ and then compute an integrating factor in $x$. That would give something like $e^{\int \frac{2y}{x+2y^3}\,dx}$, which is nonsense because $y$ is not a constant. Always check whether the roles of $x$ and $y$ can be swapped to get a linear form.

   ✓Final answer

   The correct option is (B) $\frac{1}{\sqrt{y}}$.

8. 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 $P=1$, $Q=x$.

   $\text{I.F.}=e^{\int 1\,dx}=e^x$

   ✓Final answer

   The integrating factor is $e^x$.

9. 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:

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

   Note: only P(x) (the coefficient of y) determines the integrating factor — Q(x) never enters this calculation. So even though this item's source scan had a slightly ambiguous exponent glyph in the right-hand-side term (best read as x^4), the integrating factor is unaffected either way and remains $1/x$.

   ✓Final answer

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

10. 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

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

    This is linear in $x$, of the form $\dfrac{dx}{dy}+P(y)x=Q(y)$ with $P(y)=-\dfrac1y$. The integrating factor is

    $IF = e^{\int P(y)\,dy} = e^{-\int \frac1y\,dy} = e^{-\ln y} = \frac1y.$

    ✓Final answer

    (c) $\dfrac{1}{y}$.

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.Choose the correct answer : The integrating factor of the differential equation $x\dfrac{dy}{dx} - y = 2x^2$ is

    (a) $e^{-x}$

    (b) $e^{-y}$

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

    (d) $x$

    ›Reveal solutionSolution

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

    Starting from $x\dfrac{dy}{dx}-y=2x^2$, divide throughout by $x$:

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

    This is linear in $y$ with $P(x)=-\dfrac1x$ and $Q(x)=2x$.

    The integrating factor is

    $\text{IF}=e^{\int P\,dx}=e^{\int-\frac1x\,dx}=e^{-\log x}=e^{\log x^{-1}}=\frac1x.$

    Checking against the options: (a) $e^{-x}$, (b) $e^{-y}$, (d) $x$ do not arise from $\int-\frac1x\,dx$; only (c) $\dfrac1x$ does.

    ✓Final answer

    (c) $\dfrac1x$

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.The integrating factor of differential equation cos x (dy/dx) + y sin x = 1 is .......

    (a) sec x + tan x

    (b) Log(sec x + tan x)

    (c) e^(sec x)

    (d) sec x

    ›Reveal solutionSolution

    Writing the equation in standard linear form $\dfrac{dy}{dx} + Py = Q$ gives $P = \tan x$, and the integrating factor $e^{\int P\,dx} = e^{\ln|\sec x|} = \sec x$.

    Given: $\cos x \dfrac{dy}{dx} + y\sin x = 1$

    Step 1 — standard form: Divide throughout by $\cos x$:

    $\frac{dy}{dx} + y\tan x = \sec x$

    This is linear in $y$, with $P(x) = \tan x$ and $Q(x) = \sec x$.

    Step 2 — integrating factor:

    $\text{I.F.} = e^{\int \tan x\, dx} = e^{\ln|\sec x|} = \sec x$

    ✓Final answer

    The integrating factor is $\sec x$ (Option d).