JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

y\frac{{dy}}{{dx}} = x\left[ {\frac{{{y^2}}}{{{x^2}}} + \frac{{\phi \left( {\frac{{{y^2}}}{{{x^2}}}} \right)}}{{\phi '\left( {\frac{{{y^2}}}{{{x^2}}}} \right)}}} \right],{\mkern 1mu} {\mkern 1mu} x > 0,{\mkern 1mu} {\mkern 1mu} \phi > 0

and

y(1)=-1

then

\phi \left( {\frac{{{y^2}}}{4}} \right)

is equal to

View answer and explanation

Correct option: B

4\,\,\phi \,\,(1)

Option B (

4\,\,\phi \,\,(1)

) is correct because it is the solution obtained after solving the given differential equation and applying the initial condition.

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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)