JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

Let

y=y(x)

be the solution curve of the differential equation

\frac{d y}{d x}=\frac{y}{x}\left(1+x y^{2}\left(1+\log _{e} x\right)\right), x>0, y(1)=3

. Then

\frac{y^{2}(x)}{9}

is equal to

View answer and explanation

Correct option: C

\frac{x^{2}}{5-2 x^{3}\left(2+\log _{e} x^{3}\right)}

Option C (

\frac{{x^{{2}}}}{{5-2 x^{{3}} (2+\log _{{e}} x^{{3}})}}

) is correct because it is the solution to the given differential equation with the initial condition y(1)=3.

More Questions from DIFFERENTIAL EQUATIONS

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y=y(x)

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

y(1)=1

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

x

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

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

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

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

y=y_{1}(x)

y=y_{2}(x)

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

y_{1}(0)=0

y_{2}(0)=1

y=y_{1}(x)

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)