JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 19)
Let y = y(x) be the solution of the differential equation dy = e$$\alpha$$x + y dx; $$\alpha$$ $$\in$$ N. If y(loge2) = loge2 and y(0) = loge$$\left( {{1 \over 2}} \right)$$, then the value of $$\alpha$$ is equal to _____________.
Answer
2
Explanation
$$\int {{e^{ - y}}} dy = \int {{e^{\alpha x}}} dx$$
$$ \Rightarrow {e^{ - y}} = {{{e^{\alpha x}}} \over \alpha } + c$$ ..... (i)
Put (x, y) = (ln2, ln2)
$${{ - 1} \over 2} = {{{2^\alpha }} \over \alpha } + C$$ ..... (ii)
Put (x, y) $$ \equiv $$ (0, $$-$$ln2) in (i)
$$ - 2 = {1 \over \alpha } + C$$ ..... (iii)
(ii) $$-$$ (iii)
$${{{2^\alpha } - 1} \over \alpha } = {3 \over 2}$$
$$\Rightarrow$$ $$\alpha$$ = 2 (as $$\alpha$$ $$\in$$ N)
$$ \Rightarrow {e^{ - y}} = {{{e^{\alpha x}}} \over \alpha } + c$$ ..... (i)
Put (x, y) = (ln2, ln2)
$${{ - 1} \over 2} = {{{2^\alpha }} \over \alpha } + C$$ ..... (ii)
Put (x, y) $$ \equiv $$ (0, $$-$$ln2) in (i)
$$ - 2 = {1 \over \alpha } + C$$ ..... (iii)
(ii) $$-$$ (iii)
$${{{2^\alpha } - 1} \over \alpha } = {3 \over 2}$$
$$\Rightarrow$$ $$\alpha$$ = 2 (as $$\alpha$$ $$\in$$ N)
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