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JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Statement 1: The degrees of the differential equations

\frac{d y}{d x}+y^2=x

and

\frac{d^2 y}{d x^2}+y=\sin x

are equal.

Statement 2: The degree of a differential equation, when it is a polynomial equation in derivatives, is the highest positive integral power of the highest order derivative involved in the differential equation, otherwise degree is not defined.

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Correct option: D

Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1 .

Option D (Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1) is correct because both statements are true and statement 2 correctly explains statement 1.

More Questions from DIFFERENTIAL EQUATIONS

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y=y(x)

be the solution of the differential equation

\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|

is equal to : The degree and order of the differential equation whose solution is a parabola whose axis is

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}}}

is equal to Let

y=y_{1}(x)

y=y_{2}(x)

be the solution curves of the differential equation

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

with initial conditions

y_{1}(0)=0

y_{2}(0)=1

respectively. Then the curves

y=y_{1}(x)

y=y_{2}(x)

intersect at Let

y=y(t)

be a solution of the differential equation

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)