Question

Given that dy/dx = ye^x and x = 0, y = e. Find the value of y when x = 1.

Answer 1

Answer :

Given differential equation is

dy/dx = y eˣ

⇒ (dy)/y = eˣ dx

On integration, we get

∫ (dy)/y = ∫ eˣ dx

⇒ logy = eˣ + c ...(i), where c is integral constant

When, x = 0 and y = e,

loge = e⁰ + c

⇒ 1 = 1 + c

⇒ c = 0

From (i), we get

logy = eˣ ...(ii)

When x = 1, from (ii), we get

logy = e¹

⇒ logy = e

⇒ y = eᵉ,

which is the required value of y.

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