If $y^{\prime}=\frac{x-y}{x+y}$, then its solution is

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

y^{\prime}=\frac{x-y}{x+y}

, then its solution is

View answer and explanation

Correct option: A

y^2+2 x y-x^2=c

Option A (

y^2+2 x y-x^2=c

) is correct because it represents the solution to the given differential equation after applying the substitution

y=vx

and integrating.

Practice All Chapter MCQs →

y=y(x)

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x

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

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

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y=y_{2}(x)

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y_{1}(0)=0

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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)