The solution of the differential equation ${e^{ - x}}\left( {y + 1} \right)dy + \left( {{{\cos }^2}x - \sin 2x} \right)y{\rm{ }}dx = 0$ subjected to the condition that $y = 1$ when $x = 0$ is

$\log \left( {y + 1} \right) + {e^x}{\cos ^2}x = 1$

$\left( {y + 1} \right) + {e^x}{\cos ^2}x = 2$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of the differential equation

{e^{ - x}}\left( {y + 1} \right)dy + \left( {{{\cos }^2}x - \sin 2x} \right)y\mathrm{ }dx = 0

subjected to the condition that

y = 1

when

x = 0

View answer and explanation

Correct option: A

y + \log y + {e^x}{\cos ^2}x = 2

Option A (

y + \log y + {e^x}{\cos ^2}x = 2

) is correct because it is the solution of the given differential equation that satisfies the given initial condition.

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)