JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

Let

y = y\left( x \right)

be the solution of the differential equation

x\left( {1 - {x^2}} \right)\frac{{dy}}{{dx}} + \left( {3{x^2}y - y - 4{x^3}} \right) = 0,\,x > 1,

with

y\left( 2 \right) = - 2.

Then

y\left( 3 \right)

is equal to

View answer and explanation

Correct option: A

–18

Option A (–18) is correct because it is the value of y(3) obtained by solving the given differential equation with the initial condition y(2) = -2.

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y=y(x)

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

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

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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)