JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

x^2+y^2=1

, then

\left[\right.

Note

\left.: y^{\prime}=\frac{\mathrm{dy}}{\mathrm{dx}}, \mathrm{y}^{\prime \prime}=\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right]

View answer and explanation

Correct option: B

y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0

Option B (

y y^{\prime \prime}+(y^{\prime})^2+1=0

) is correct because it is the result of differentiating the given equation twice and eliminating x.

More Questions from DIFFERENTIAL EQUATIONS

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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

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

x

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

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

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

\frac{{{d^2}x}}{{d{t^2}}}

y=y_{1}(x)

y=y_{2}(x)

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

y_{1}(0)=0

y_{2}(0)=1

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)