JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Hard

1 of 1

Application +4 | -1

y'' - 3y' + 2y = 0

where

y\left( 0 \right) = 1,y'\left( 0 \right) = 0

, then the value of

y

x = \log 2

View answer and explanation

Correct option: D

Option D (0) is correct because substituting

x = \log 2

into the solution of the differential equation, obtained using the given initial conditions, yields

y = 0

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

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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)