Q.Integrating factor of $x\frac{dy}{dx}-y=x^4-3x$ is:
(A) $x$
(B) $\log x$
(C) $\frac{1}{x}$
(D) $-x$

Showing the 12 most recent of 33 on this concept.

1. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.An integrating factor of the differential equation $(x^2+1)\dfrac{dy}{dx} + xy = x^3$ is (A) $\dfrac{x}{1+x^2}$ (B) $\dfrac{1}{2}\log(1+x^2)$ (C) $\sqrt{1+x^2}$ (D) $e^{\log(1+x^2)}$

   ›Reveal solutionSolution

   Writing the ODE in standard linear form $\frac{dy}{dx}+Py=Q$ and computing $e^{\int P\,dx}$ gives the integrating factor $\sqrt{1+x^2}$.

   Concept and Intuition

   A first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$ is solved by multiplying through by the integrating factor $\mu(x)=e^{\int P(x)\,dx}$, chosen precisely so that $\dfrac{d}{dx}\big[\mu(x)y\big]=\mu(x)Q(x)$. So the entire method hinges on correctly identifying $P(x)$ after dividing the equation into standard form.

   Step-by-Step Solution

   1. Start with $(x^2+1)\dfrac{dy}{dx}+xy=x^3$.
   2. Divide by $(x^2+1)$ to reach standard form: $\dfrac{dy}{dx}+\dfrac{x}{x^2+1}y=\dfrac{x^3}{x^2+1}$.
   3. Here $P(x)=\dfrac{x}{x^2+1}$.
   4. Compute $\int P(x)\,dx=\displaystyle\int \dfrac{x}{x^2+1}dx=\dfrac12\log(x^2+1)$ (substituting $u=x^2+1$). …

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2. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The integrating factor of the linear differential equation $\dfrac{dy}{dx} = \dfrac{1}{4x+3y}$ is (A) $e^{-4x}$ (B) $e^{4x}$ (C) $e^{3y}$ (D) $e^{-3y}$

   ›Reveal solutionSolution

   The ODE is not linear in $y$, but writing $x$ as the dependent variable turns it into $\dfrac{dx}{dy}-4x=3y$, whose integrating factor is $e^{-4y}$ — the option carrying the exponent $-4$.

   Concept and Intuition

   A first-order linear ODE must look like $\dfrac{dy}{dx}+P(x)y=Q(x)$. Here $\dfrac{dy}{dx}=\dfrac{1}{4x+3y}$ has the unknown buried in a denominator, so it is not linear in $y$. The rescue is a change of viewpoint: nothing stops us from regarding $x$ as the function and $y$ as the independent variable. Since $\dfrac{dx}{dy}=1\big/\dfrac{dy}{dx}$, the reciprocal instantly clears the denominator and the equation becomes linear in $x$. The integrating factor is then $e^{\int P\,dy}$, with $P$ the coefficient of $x$.

   Step-by-Step Solution

   1. Invert: $\dfrac{dx}{dy}=4x+3y$.
   2. Put it in standard linear form (dependent variable $x$):

   $\frac{dx}{dy}-4x=3y,\qquad P(y)=-4,\ \ Q(y)=3y .$

   3. Integrating factor: $\text{I.F.}=e^{\int P\,dy}=e^{\int(-4)\,dy}=e^{-4y}.$ …

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3. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\log y$ is an integrating factor of $\dfrac{dx}{dy}+P(y)x=Q(y)$, then $P(y)=$ (A) $\dfrac{1}{y+\log y}$ (B) $\dfrac{y}{\log y}$ (C) $\dfrac{\log y}{y}$ (D) $\dfrac{1}{y\log y}$

   ›Reveal solutionSolution

   Working backward from the known integrating factor $\log y=e^{\int P(y)dy}$ and differentiating recovers $P(y)=\dfrac{1}{y\log y}$.

   Concept and Intuition

   For a linear ODE $\dfrac{dx}{dy}+P(y)x=Q(y)$ (linear in $x$, with $y$ as the independent variable), the integrating factor is $\mathrm{IF}=e^{\int P(y)\,dy}$. If we're told what the integrating factor equals, we can reverse-engineer $P(y)$ by taking logs and then differentiating.

   Step-by-Step Solution

   1. Standard fact: $\mathrm{IF}=e^{\int P(y)\,dy}$.
   2. Given $\mathrm{IF}=\log y$ (i.e. $\log y$), set $e^{\int P(y)dy}=\log y$.
   3. Take the natural log of both sides: $\displaystyle\int P(y)\,dy=\log(\log y)$.
   4. Differentiate both sides with respect to $y$ (by the Fundamental Theorem of Calculus, the left side gives back $P(y)$): $P(y)=\dfrac{d}{dy}\left[\log(\log y)\right]$.
   5. By the chain rule: $\dfrac{d}{dy}\log(\log y)=\dfrac{1}{\log y}\cdot\dfrac{d}{dy}(\log y)=\dfrac{1}{\log y}\cdot\dfrac1y=\dfrac{1}{y\log y}$. …

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4. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Integrating factor of the differential equation $\sin x \dfrac{dy}{dx} - y\cos x = 1$ is (A) $\sin x$ (B) $\cos x$ (C) $\sec x$ (D) $\mathrm{cosec}\, x$

   ›Reveal solutionSolution

   Rewriting the equation in standard linear form and computing $e^{\int P\,dx}$ gives the integrating factor $\csc x$ — (D).

   Concept and Intuition

   A first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$ has integrating factor $e^{\int P(x)dx}$. The given equation must first be divided through so the coefficient of $\dfrac{dy}{dx}$ is exactly $1$.

   Step-by-Step Solution

   1. Given: $\sin x\dfrac{dy}{dx}-y\cos x=1$.
   2. Divide by $\sin x$: $\dfrac{dy}{dx}-\dfrac{\cos x}{\sin x}y=\dfrac1{\sin x}$, i.e. $\dfrac{dy}{dx}-(\cot x)y=\csc x$.
   3. Here $P(x)=-\cot x$.
   4. Integrating factor $=e^{\int -\cot x\,dx}=e^{-\log|\sin x|}=(\sin x)^{-1}=\csc x$.

   Common Mistakes …

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5. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.The solution of differential equation $(x + 2y^3)\dfrac{dy}{dx} = y$ is (A) $x = y(2xy + c)$ (B) $x = y(y^2 + c)$ (C) $y = x(x^2 + c)$ (D) $xy = \dfrac{y^4}{2} + c$

   ›Reveal solutionSolution

   Flipping the roles of $x$ and $y$ turns this into a standard linear ODE in $x$; solving it gives $x = y(y^2+c)$.

   Concept and Intuition

   The equation $(x+2y^3)\dfrac{dy}{dx}=y$ is not linear in $y$, but if we invert it and treat $x$ as a function of $y$, it becomes linear in $x$ — a very common trick when the equation is "linear except the dependent/independent variables are swapped."

   Step-by-Step Solution

   1. Invert: $\dfrac{dx}{dy}=\dfrac{x+2y^3}{y}$.
   2. Rearrange: $\dfrac{dx}{dy}-\dfrac{1}{y}x = 2y^2$. This is linear in $x$ with $P(y)=-\dfrac{1}{y}$, $Q(y)=2y^2$.
   3. Integrating factor: $IF=e^{\int P\,dy}=e^{-\int dy/y}=e^{-\log y}=\dfrac{1}{y}$.
   4. Multiply through: $\dfrac{d}{dy}\!\left(\dfrac{x}{y}\right)=\dfrac{2y^2}{y}=2y$. …

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6. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.The integrating factor of the linear differential equation $\dfrac{dy}{dx} + P(x)y = Q(x)$ is a solution of the differential equation (A) $\dfrac{dy}{dx} - P(x)y = 0$ (B) $\dfrac{dy}{dx} + P(x)y = 0$ (C) $\dfrac{dy}{dx} - \dfrac{y}{x} = P(x)$ (D) $\dfrac{dy}{dx} + \dfrac{x}{y} = P(x)$

   ›Reveal solutionSolution

   The integrating factor $\mu=e^{\int P\,dx}$ is itself a solution of the homogeneous equation $y'=P(x)y$.

   Concept and Intuition

   The whole point of the integrating factor is that $\dfrac{d}{dx}\big(\mu y\big)=\mu\dfrac{dy}{dx}+\mu P(x)y$ should equal $\mu\left(\dfrac{dy}{dx}+P(x)y\right)$, which requires $\dfrac{d\mu}{dx}=\mu P(x)$ — i.e. $\mu$ itself solves the first-order linear homogeneous ODE $y'-P(x)y=0$.

   Step-by-Step Solution

   1. The integrating factor is defined as $\mu(x)=e^{\int P(x)\,dx}$.
   2. Differentiate: $\mu'(x) = P(x)\,e^{\int P\,dx} = P(x)\mu(x)$.
   3. Rearranged: $\mu'(x) - P(x)\mu(x)=0$. …

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7. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx}+\dfrac{y}{x}=x^2$ is (A) $y=\dfrac{1}{3}x^3+\dfrac{c}{x}$ (B) $y=\dfrac{1}{4}x^4+cx$ (C) $y=\dfrac{1}{4}x^3+c$ (D) $y=\dfrac{1}{4}x^3+cx^{-1}$

   ›Reveal solutionSolution

   A standard first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$; the integrating factor $x$ makes the left side an exact derivative, giving the general solution $y=\dfrac14x^3+\dfrac{c}{x}$.

   Concept and Intuition

   For a linear equation $\dfrac{dy}{dx}+P(x)y=Q(x)$, multiplying both sides by the integrating factor $\mu(x)=e^{\int P\,dx}$ turns the left side into the exact derivative $\dfrac{d}{dx}\big(\mu y\big)$, which can then be integrated directly.

   Step-by-Step Solution

   1. Here $P(x)=\dfrac1x$, $Q(x)=x^2$.
   2. Integrating factor: $\mu(x)=e^{\int \frac1x dx}=e^{\log x}=x$.
   3. Multiply through: $x\dfrac{dy}{dx}+y=x^3$, i.e. $\dfrac{d}{dx}(xy)=x^3$.
   4. Integrate both sides: $xy=\displaystyle\int x^3\,dx=\dfrac{x^4}{4}+c$. …

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8. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The general solution of the equation $\dfrac{dy}{dx}+\dfrac{1}{x}y=\dfrac{1}{x}e^x$ is (A) $y=xe^x+c$ (B) $y=xe^x+ce^{-x}$ (C) $y=\dfrac{e^x+c}{x}$ (D) $y=\dfrac{e^{-x}+cx}{x}$

   ›Reveal solutionSolution

   This is a standard linear first-order ODE solved via an integrating factor; the answer is (C).

   Concept and Intuition

   An equation of the form $\dfrac{dy}{dx}+P(x)y=Q(x)$ is solved by multiplying through by the integrating factor $\mu=e^{\int P\,dx}$, which makes the left side a perfect derivative $\dfrac{d}{dx}(\mu y)$.

   Step-by-Step Solution

   1. Here $P(x)=\dfrac1x$, $Q(x)=\dfrac1x e^x$.
   2. Integrating factor: $\mu=e^{\int \frac1x dx}=e^{\log x}=x$.
   3. Multiply the ODE by $x$: $x\dfrac{dy}{dx}+y=e^x$, i.e. $\dfrac{d}{dx}(xy)=e^x$.
   4. Integrate both sides: $xy=e^x+c$. …

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9. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If $x\,dy+(y+y^2x)\,dx=0$ and $y=1$ at $x=1$, then (A) $y=\dfrac{x}{1+\log x}$ (B) $y=\dfrac{1+\log x}{x}$ (C) $y=x(1+\log x)$ (D) $y=\dfrac{1}{x(1+\log x)}$

   ›Reveal solutionSolution

   A Bernoulli equation in disguise; the substitution $v=1/y$ linearizes it, and the initial condition $y(1)=1$ pins down the constant to give $y=\dfrac{1}{x(1+\log x)}$.

   Concept and Intuition

   An equation of the form $\dfrac{dy}{dx}+P(x)y=Q(x)y^n$ is a Bernoulli equation. Dividing by $y^n$ and substituting $v=y^{1-n}$ converts it into a linear first-order ODE in $v$, which can then be solved with the standard integrating-factor method.

   Step-by-Step Solution

   1. Given $x\,dy+(y+y^2x)\,dx=0$. Divide by $dx$: $x\dfrac{dy}{dx}+y+y^2x=0$.
   2. Divide by $x$: $\dfrac{dy}{dx}+\dfrac{y}{x}=-y^2$ — a Bernoulli equation with $n=2$.
   3. Divide throughout by $y^2$: $y^{-2}\dfrac{dy}{dx}+\dfrac{1}{xy}=-1$.
   4. Let $v=y^{-1}$, so $\dfrac{dv}{dx}=-y^{-2}\dfrac{dy}{dx}$, i.e. $y^{-2}\dfrac{dy}{dx}=-\dfrac{dv}{dx}$.
   5. Substitute: $-\dfrac{dv}{dx}+\dfrac{v}{x}=-1 \Rightarrow \dfrac{dv}{dx}-\dfrac{v}{x}=1$, a linear ODE in $v$.
   6. Integrating factor $=e^{\int -1/x\,dx}=e^{-\log x}=\dfrac{1}{x}$.
   7. $\dfrac{d}{dx}\!\left(\dfrac{v}{x}\right)=\dfrac{1}{x} \Rightarrow \dfrac{v}{x}=\log x+C \Rightarrow v=x(\log x+C)$. …

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10. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.By multiplying with $e^{\int P\,dx}$ on both sides of the equation $\dfrac{dy}{dx} + P(x)y = Q(x)$, the left side of the equation takes the form $\dfrac{d}{dx}(y\,f(x))$, then $f(x) =$ (A) $\displaystyle\int y\,e^{\int P\,dx}\,dx$ (B) $y\,P(x)$ (C) $e^{\int P\,dx}$ (D) $P(x)\,e^{\int P\,dx}$

    ›Reveal solutionSolution

    This is the definition-check behind the standard integrating-factor method for first-order linear ODEs: the multiplier that turns the left side into an exact derivative $\frac{d}{dx}(y\cdot(\text{something}))$ is, by construction, the integrating factor itself.

    Concept and Intuition

    The whole point of the integrating factor $\mu(x)=e^{\int P\,dx}$ is that $\dfrac{d\mu}{dx}=P(x)\mu(x)$ (by the fundamental theorem of calculus applied to the exponent). This special property is exactly what's needed to make $\mu\dfrac{dy}{dx}+\mu P y$ collapse into a single product-rule derivative.

    Step-by-Step Solution

    1. Start with $\dfrac{dy}{dx}+P(x)y=Q(x)$ and multiply both sides by $\mu(x)=e^{\int P\,dx}$:

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

    2. Compute $\dfrac{d\mu}{dx}$: since $\mu=e^{\int P\,dx}$, by the chain rule $\dfrac{d\mu}{dx}=P(x)\cdot e^{\int P\,dx}=P(x)\mu$. …

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11. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.The general solution of the differential equation $x\log x\,dy = (x\log x - y)\,dx$ is (A) $(x - y)\log x + x = c$ (B) $x - y = \dfrac{x}{\log x} + c$ (C) $y - x = \dfrac{x}{\log x} + c$ (D) $(y - x)\log x + x = c$

    ›Reveal solutionSolution

    This tests recognizing a first-order LINEAR differential equation in disguise and applying the standard integrating-factor method.

    Concept and Intuition

    After dividing through by $x\log x$, the equation takes the standard linear form $\dfrac{dy}{dx}+P(x)y=Q(x)$ with $P(x)=\dfrac{1}{x\log x}$. The integrating factor $e^{\int P\,dx}$ makes the left side an exact derivative $\dfrac{d}{dx}(y\cdot \text{IF})$, so the whole equation integrates directly.

    Step-by-Step Solution

    1. Divide both sides by $x\log x$: $\dfrac{dy}{dx}=1-\dfrac{y}{x\log x}\ \Rightarrow\ \dfrac{dy}{dx}+\dfrac{y}{x\log x}=1$.
    2. Integrating factor: $\mathrm{IF}=\exp\!\left(\int \dfrac{dx}{x\log x}\right)=\exp(\log(\log x))=\log x$ (since $\int\frac{dx}{x\log x}=\log(\log x)+C$).
    3. Multiply the ODE by $\log x$: $\dfrac{d}{dx}\big(y\log x\big)=\log x$. …

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12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.The general solution of the differential equation $(x + 2y^3)\dfrac{dy}{dx} - y = 0$, $y > 0$ is (A) $y = x^3 + cy$ (B) $x = y^3 + cy$ (C) $y(1 - xy) = cx$ (D) $x(1 - xy) = cy$

    ›Reveal solutionSolution

    Treat $x$ as the dependent variable and $y$ as independent — the equation is linear in $x$ once rearranged. Answer: $x=y^3+cy$.

    Concept and Intuition

    The equation $(x+2y^3)\dfrac{dy}{dx}-y=0$ is not linear in $y$ (because of the $y^3$ term), but if we instead regard $x$ as a function of $y$, it becomes linear in $x$. This is a standard trick: whenever an ODE is linear in one variable when viewed "the other way around," solve it as $\dfrac{dx}{dy}+P(y)x=Q(y)$ instead of $\dfrac{dy}{dx}+P(x)y=Q(x)$.

    Step-by-Step Solution

    1. Given: $(x+2y^3)\dfrac{dy}{dx} = y \implies \dfrac{dx}{dy} = \dfrac{x+2y^3}{y}$ (inverting the derivative, valid since $y>0$).
    2. Expand: $\dfrac{dx}{dy} = \dfrac{x}{y} + 2y^2 \implies \dfrac{dx}{dy} - \dfrac{1}{y}x = 2y^2$.
    3. This is linear in $x$ with $P(y)=-\dfrac1y$, $Q(y)=2y^2$.
    4. Integrating factor: $\mu(y) = e^{\int -\frac1y dy} = e^{-\log y} = \dfrac1y$.
    5. Then $\dfrac{d}{dy}\!\left(\dfrac{x}{y}\right) = \dfrac1y\cdot 2y^2 = 2y$.
    6. Integrate: $\dfrac{x}{y} = \displaystyle\int 2y\,dy = y^2 + c$. …

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