JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

Let us consider a curve,

y = f(x)

passing through the point

( - 2,\,\,2)

and the slope of the tangent to the curve at any point

(x,f(x))

is given by

f(x) + xf'(x)\, = {x^2}

then

View answer and explanation

Correct option: C

{x^3} - 3x\mathrm{ }f(x) - 4 = 0

The correct answer is

{x^3} - 3x\mathrm{ }f(x) - 4 = 0

because it is obtained by solving the given differential equation and using the given point to find the constant of integration.

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)