Let y=yx be the solution of the differential equation dy=e^(αx+y)dx;α∈N. If ylog_e2=log_e2 and y0=log_e1/2, then the value of α is equal to ___.

Correct answer: 2. We have, ∫e^(-y)dy=∫e^(αx)dx ⇒-e^(-y)=e^(αx)α+c ...1 Put x,y=ℓn2,ℓn2 -1/2=2^(α)α+c ...2 Put x,y≡0,-ℓn2 -2=1/(α)+c From equation 1-2, we get (2^(α)-1)/(α)=3/2 ⇒α=2( as α∈N).

Q56 / 90 · JEE Main 2021 (27 Jul Shift 2) · Mathematics

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N .$ If $y (\log_{e} 2) = \log_{e} 2$ and $y (0) = \log_{e} (\frac{1}{2}),$ then the value of $α$ is equal to ___.

Correct answer: 2

Solution

We have, $\int e^{- y} d y = \int e^{α x} d x$

$\Rightarrow - e^{- y} = \frac{e^{α x}}{α} + c . . . (1)$

Put $(x, y) = (ℓ n 2, ℓ n 2)$

$- \frac{1}{2} = \frac{2^{α}}{α} + c . . . (2)$

Put $(x, y) \equiv (0, - ℓ n 2)$

$- 2 = \frac{1}{α} + c$

From equation $(1) - (2),$ we get

$\frac{2^{α} - 1}{α} = \frac{3}{2}$

$\Rightarrow α = 2 ($ as $α \in N)$ .

Hence, the correct answer is 2.

Let y=yx be the solution of the differential equation dy=eαx+ydx;α∈N. If yloge2=loge2 and y0=loge12, then the value of α is equal to ___.

Solution

Relevant Questions

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N .$ If $y (\log_{e} 2) = \log_{e} 2$ and $y (0) = \log_{e} (\frac{1}{2}),$ then the value of $α$ is equal to ___.