$\frac{d y}{d x}+y=2 e^{2 x}$ then $y$ is

$(1+\mathrm{x}) \mathrm{e}^{-\mathrm{x}}+\frac{2}{3} \mathrm{e}^{2 \mathrm{x}}+\mathrm{c}$

$\mathrm{ce}^{-\mathrm{x}}+\frac{2}{3} \mathrm{e}^{2 \mathrm{x}}+\mathrm{c}$

$\mathrm{e}^{-\mathrm{x}}+\frac{2}{3} \mathrm{e}^{2 \mathrm{x}}+\mathrm{c}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

\frac{d y}{d x}+y=2 e^{2 x}

then

y

View answer and explanation

Correct option: A

\operatorname{ce}^{-\mathrm{x}}+\frac{2}{3} \mathrm{e}^{2 \mathrm{x}}

Option A (

ce^{-\mathrm{x}}+\frac{{2}}{{3}} \mathrm{e}^{2 \mathrm{x}}

) is correct because it is the general solution to the given first-order linear differential equation.

Practice All Chapter MCQs →

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x

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

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y=y_{2}(x)

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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)