JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

Let

y = \,y(x)

be solution of the differential equation

{\log _e}\left( {\frac{{dy}}{{dx}}} \right) = 3x + 4y

, with

y(0) = \,0

.
If

y\left( { - \frac{2}{3}{{\log }_e}2} \right) = \alpha {\log _e}2

, then the value of

\alpha

is equal to

View answer and explanation

Correct option: A

- \frac{1}{4}

Option A (-1/4) is correct because solving the differential equation and using the given condition y(0) = 0, we can find the value of alpha.

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)