JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

y = f\left( x \right)

is the equation of the curve an its differential equation is given by

\frac{{dy}}{{dx}} = \frac{{x + 2}}{{y + 3}},

then the equation of the curve, if it passes through (2,2), is

View answer and explanation

Correct option: A

{x^2} - {y^2} + 4x - 6y + 4 = 0

Option A (

{x^2} - {y^2} + 4x - 6y + 4 = 0

) is correct because it satisfies the given differential equation and passes through the point (2,2).

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)