Question

Find the explicit solution to the initial value problem (eʸ²/2y )+eˣ/ˣ⁺¹ dy/dx = 0 where y(0) = √(ln2).

Answer 1

To solve the initial value problem (eʸ²/2y )+eˣ/ˣ⁺¹ dy/dx = 0 where y(0) = √(ln2), we can separate the variables, integrate, and use the initial condition to find the explicit solution y² = -eˣ + 2 - ln2.

Step-by-step explanation:

To solve the initial value problem (eʸ²/2y )+eˣ/ˣ⁺¹ dy/dx = 0 where y(0) = √(ln2),

we can separate the variables and integrate.

Multiplying both sides of the equation by 2y/eʸ² and by ˣ⁺¹, we get:

2ydy = -eˣ dx.

Integrating both sides, we have:

∫2ydy = -∫eˣ dx.

Integrating, we get:

y² = -eˣ + C.

Using the initial condition y(0) = √(ln2), we can solve for C:

(√(ln2))² = -e⁰ + C.

Simplifying, we find:

C = 2-ln2.

So, the explicit solution to the initial value problem is:

y² = -eˣ + 2 - ln2.