JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The solution of

\frac{x \mathrm{~d} y}{x^2+y^2}=\left(\frac{y}{x^2+y^2}-1\right)

\mathrm{d} x

View answer and explanation

Correct option: C

y=x \tan (c-x)

Option C (

y=x \tan (c-x)

) is correct because it is the solution obtained by rewriting the given differential equation, separating variables, and integrating.

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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

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

x

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

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

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

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y=y_{2}(x)

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

y_{1}(0)=0

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)