Q.Find the general solution of the differential equation $y e^y dx = (y^3 + 2x e^y) dy$.

Showing the 12 most recent of 41 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 $\text{IF} = e^{\int \frac{P}{R}\, dy}.$ …

   🔒 Start your 14-day free trial to see the full solution →
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 $Q(x)$:

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

   $$ Q(x) = \frac{3}{2x} $$ …
   

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set 65/3/11 markMCQ
   Q.$\frac{dy}{dx} = F(x, y)$ will be a homogeneous differential equation for which of the following functions?

   (i) $F(x, y) = 3x + 2y$

   (ii) $F(x, y) = \sin \frac{y}{x} + \log y - \log x$

   (iii) $F(x, y) = e^{y/x} + 1$

   (iv) $F(x, y) = \sqrt{x^2 + y^2} - y$ (A)

   (i) and

   (ii) (B) (i),

   (ii) and

   (iii) (C) (ii),

   (iii) and

   (iv) (D)

   (ii) and (iii)

   ›Reveal solutionSolution

   A differential equation $\frac{dy}{dx} = F(x, y)$ is homogeneous exactly when $F$ is homogeneous of degree zero — meaning $F(tx, ty) = F(x, y)$ for all $t > 0$, which is equivalent to $F$ being expressible purely as a function of $y/x$. Only options (ii) and (iii) satisfy this, so the correct choice is (D).

   The idea is simple: a homogeneous differential equation is one where the right-hand side $F(x, y)$ doesn't change if you scale both $x$ and $y$ by the same factor. Why does that matter? Because if $F$ has that property, you can substitute $y = vx$ and turn the equation into one in $v$ and $x$ alone — a separable equation you can actually solve. The test is clean: check whether $F(tx, ty) = F(x, y)$ for any $t > 0$.

   Let's go through each option.

   1. Option (i): $F(x, y) = 3x + 2y$ Replace $x$ with $tx$ and $y$ with $ty$:

   $F(tx, ty) = 3(tx) + 2(ty) = t(3x + 2y) = t \cdot F(x, y)$

   This is $t$ times the original, not equal to it — unless $t = 1$. So $F$ is homogeneous of degree 1, not degree 0. It also cannot be written as a function of $y/x$ alone (try it: $3x + 2y = x(3 + 2(y/x))$ still has an $x$ factor outside). So this is not homogeneous for the purpose of $\frac{dy}{dx} = F(x, y)$.

   2. Option (ii): $F(x, y) = \sin\frac{y}{x} + \log y - \log x$ First simplify the log terms: $\log y - \log x = \log\frac{y}{x}$. So

   $F(x, y) = \sin\frac{y}{x} + \log\frac{y}{x}$

   This is already written purely in terms of $y/x$. That's a dead giveaway — it's homogeneous of degree 0. Check formally:

   $F(tx, ty) = \sin\frac{ty}{tx} + \log\frac{ty}{tx} = \sin\frac{y}{x} + \log\frac{y}{x} = F(x, y)$

   The $t$ cancels completely. So this is homogeneous.

   3. Option (iii): $F(x, y) = e^{y/x} + 1$ Again, this is already a function of $y/x$ alone. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $P=\dfrac{1}{1+x^2}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $u = 1-y^2$, $du = -2y\,dy$: …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 $P=1/x$.

   …

   🔒 Start your 14-day free trial to see the full solution →
7. 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

   …

   🔒 Start your 14-day free trial to see the full solution →
8. 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

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

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $P=1$, $Q=x$.

   …

   🔒 Start your 14-day free trial to see the full solution →
10. 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:

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

    …

    🔒 Start your 14-day free trial to see the full solution →
11. 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

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

    …

    🔒 Start your 14-day free trial to see the full solution →
12. 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

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

    …

    🔒 Start your 14-day free trial to see the full solution →