JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

y\left( t \right)

is a solution of

\left( {1 + t} \right)\frac{{dy}}{{dt}} - ty = 1\,

and

y\left( 0 \right) = - 1

, then

y\left( 1 \right)

is equal to

View answer and explanation

Correct option: A

- \frac{1}{2}

Option A (

\frac{{ - 1}}{2}

) is correct because it is the value of y(1) obtained by solving the given differential equation and using the initial condition.

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{x^2} + {y^2} = 1

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x = A\cos 4t + B\sin 4t,

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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)