The solution of the equation $\frac{{dy}}{{dx}} = \frac{{x + y}}{{x - y}}$ is

$c\left( {{x^2} - {y^2}} \right) = {e^{{{\tan }^{ - 1}}\left( {\frac{y}{x}} \right)}}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The solution of the equation

\frac{{dy}}{{dx}} = \frac{{x + y}}{{x - y}}

View answer and explanation

Correct option: B

c{\left( {{x^2} + {y^2}} \right)^{\frac{1}{2}}} = {e^{{{\tan }^{ - 1}}\left( {\frac{y}{x}} \right)}}

Option B (

c{\left( {{x^2} + {y^2}} \right)^{\frac{1}{2}}} = {e^{{{\tan }^{ - 1}}\left( {\frac{y}{x}} \right)}}

) is correct because it is the solution obtained by solving the homogeneous differential equation using the substitution y = vx.

Practice All Chapter MCQs →

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

\alpha>0, \beta>0

\gamma>0

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