If integrating factor of $x\left( {1 - {x^2}} \right)dy + \left( {2{x^2}y - y - a{x^3}} \right)dx = 0$ is ${e^{\smallint Pdx}}$, then $P$ is equal to

$\frac{{2{x^2} - a{x^3}}}{{x\left( {1 - {x^2}} \right)}}$

$\frac{{2{x^2} - 1}}{{a{x^3}}}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

If integrating factor of

x\left( {1 - {x^2}} \right)dy + \left( {2{x^2}y - y - a{x^3}} \right)dx = 0

{e^{\int Pdx}}

, then

P

is equal to

View answer and explanation

Correct option: D

\frac{{2{x^2} - 1}}{{x\left( {1 - {x^2}} \right)}}

Option D (

\frac{{2{x^2} - 1}}{{x\left( {1 - {x^2}} \right)}}

) is correct because it represents the coefficient of

y

when the given differential equation is written in the standard form

\frac{{dy}}{{dx}} + Py = Q

Practice All Chapter MCQs →

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x

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x = A\cos 4t + B\sin 4t,

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y=y_{2}(x)

\frac{d y}{d x}=y+7

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y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)