The solution of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 2x}}$ is $y = {c_1}{e^{ - 2x}} + {c_2}x + {x_3}$, where ${c_1}$ is

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Medium

1 of 1

Application +4 | -1

The solution of the differential equation

\frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 2x}}

y = {c_1}{e^{ - 2x}} + {c_2}x + {x_3}

, where

{c_1}

View answer and explanation

Correct option: B

\frac{1}{4}

Option B (

\frac{1}{4}

) is correct because it is the coefficient of the exponential term in the particular solution obtained by integrating the given differential equation twice.

More Questions from DIFFERENTIAL EQUATIONS

Practice All Chapter MCQs →

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)