The solution of $(y+x+5) \mathrm{d} y=(y-x+$ 1) $\mathrm{d} x$ is

$\log \left((y+3)^2+(x+2)^2\right)+\tan ^{-1} \frac{y+3}{x+2}=C$

$\log \left((y+3)^2+(x-2)^2\right)+\tan ^{-1} \frac{y-3}{x-2}=C$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of

(y+x+5) \mathrm{d} y=(y-x+

\mathrm{d} x

View answer and explanation

Correct option: C

\log \left((y+3)^2+(x+2)^2\right)+2 \tan ^{-1} \frac{y+3}{x+2}=C

The correct answer is C (

\log \left((y+3)^2+(x+2)^2\right)+2 \tan ^{-1} \frac{{y+3}}{{x+2}}=C

) because it is the solution to the homogeneous differential equation after substitution and integration.

Practice All Chapter MCQs →

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)