JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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The solution of the differential equation

\frac{d y}{d x}+\frac{x y}{1-x^2}=x \sqrt{y},(|x|<1)

\sqrt{y}=-\frac{f(x)}{3}+C\left(1-x^2\right)^{\frac{1}{4}}

, where

f\left(\frac{1}{2}\right)=\frac{3}{4}

and

C

is an arbitrary constant. Then, the value of

f\left(-\frac{1}{2}\right)

View answer and explanation

Correct option: B

\frac{3}{4}

Option B (

\frac{3}{4}

) is correct because it is the value of the function

f(x)

x=-\frac{1}{2}

, where

f(x)

is derived from solving the given differential equation.

More Questions from DIFFERENTIAL EQUATIONS

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y=y(x)

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y_{2}(0)=1

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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)