JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Let

x=x(t)

and

y=y(t)

be solutions of the differential equations

\frac{d x}{d t}+a x=0

and

\frac{d y}{d t}+b y=0

respectively,

a, b \in \mathbb{R}.

Given that

x(0)=2 ; y(0)=1

and

3 y(1)=2 x(1),

the value of

t,

for which

x(t)=y(t),

is:

View answer and explanation

Correct option: C

\log _{\frac{4}{3}} 2

Option C is correct because it involves solving two first-order differential equations and then using the given condition 3y(1) = 2x(1) to find the value of t for which x(t) = y(t).

More Questions from DIFFERENTIAL EQUATIONS

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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)