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

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

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\mathrm{\,\, is \,\,}{e^{\int P dx}}

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(1-x^2)}

) is correct because it represents the coefficient of 'y' in the standard form of the given linear differential equation, which is required to find the integrating factor.

More Questions from DIFFERENTIAL EQUATIONS

Practice All Chapter MCQs →

Let

y=y(x)

be the solution of the differential equation

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

such that

y(1)=1

. Then

\left|(y(2))^{3}-12 y(2)\right|

is equal to : The degree and order of the differential equation whose solution is a parabola whose axis is

x

-axis, are If

{x^2} + {y^2} = 1

, then

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

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

then

\frac{{{d^2}x}}{{d{t^2}}}

is equal to Let

y=y_{1}(x)

and

y=y_{2}(x)

be the solution curves of the differential equation

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

with initial conditions

y_{1}(0)=0

and

y_{2}(0)=1

respectively. Then the curves

y=y_{1}(x)

and

y=y_{2}(x)

intersect at Let

y=y(t)

be a solution of the differential equation

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

where,

\alpha>0, \beta>0

and

\gamma>0

. Then

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