JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Hard

1 of 1

Analyzing +4 | -1

The solution of the differential equation

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

View answer and explanation

Correct option: A

\left| {\frac{x}{y}} \right| + \frac{{xy}}{{\left( {x - y} \right)}} = c

Option A (In

\left| {\frac{x}{y}} \right| + \frac{{xy}}{{\left( {x - y} \right)}} = c

) is correct because it represents the solution obtained by integrating the given differential equation.

More Questions from DIFFERENTIAL EQUATIONS

Practice All Chapter MCQs →

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)