Let $y=y(x)$ be solution of the differential equation $\log _{e}\left(\frac{d y}{d x}\right)=3 x+4 y$, with $y(0)=0$. If $y\left(-\frac{2}{3} \log _{e} 2\right)=\alpha \log _{e} 2$, then the value of $\alpha$ is equal to :

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Differential Equations

Let

y = y (x)

be solution of the differential equation

lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y

, with

y (0) = 0

. If

y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2

, then the value of

α

is equal to :

Answer: (B)

The equation gives

\frac{d y}{d x} = e^{3 x + 4 y} = e^{3 x} e^{4 y}

, so

e^{- 4 y} d y = e^{3 x} d x \Rightarrow - \frac{e ^{- 4 y}}{4} = \frac{e ^{3 x}}{3} + c

With

y (0) = 0

- \frac{1}{4} = \frac{1}{3} + c

, so

c = - \frac{7}{12}

and

e^{- 4 y} = \frac{7 - 4 e ^{3 x}}{3}

x = - \frac{2}{3} lo g_{e} 2

e^{3 x} = e^{- 2 l o g_{e} 2} = \frac{1}{4}

, so

e^{- 4 y} = \frac{7 - 1}{3} = 2

, giving

- 4 y = lo g_{e} 2

and

y = - \frac{1}{4} lo g_{e} 2

. Hence

α = - \frac{1}{4}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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