JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

y(t)

is a solution of

(1+t) \frac{d y}{d t}-t y=1

and

y(0)=1

, then

y(1)

is equal to

View answer and explanation

Correct option: A

-\frac{1}{2}

Option A (

y=x \sec x

) is correct because it is the solution of the given differential equation with the initial condition y(0) = 0.

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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

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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_{2}(x)

\frac{d y}{d x}=y+7

y_{1}(0)=0

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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)