Find 𝑦′′(0), if 𝑦(𝑥) is implicitly defined by 𝑥5 + 4𝑦 = 𝑥 + 𝑦5 + 3, and 𝑦(0) = 1.

To find 𝑦′′(0), we need to differentiate the given equation twice with respect to 𝑥 and then evaluate the resulting expression at 𝑥 = 0, using 𝑦(0) = 1.

First, we differentiate the given equation with respect to 𝑥 to obtain:

5𝑥4 + 4𝑦′ = 1 + 5𝑦4𝑦′

Next, we differentiate this expression again with respect to 𝑥 to obtain:

20𝑥3 + 4𝑦′′ = 20𝑦4(𝑦′)2 + 5𝑦4𝑦′′ + 20𝑦3𝑦′′

Now we can substitute 𝑥 = 0 and 𝑦(0) = 1 into these expressions to get:

5(0)4 + 4𝑦′(0) = 1 + 5(1)4𝑦′(0) (evaluated at 𝑥 = 0)

20(0)3 + 4𝑦′′(0) = 20(1)4(𝑦′(0))2 + 5(1)4𝑦′′(0) + 20(1)3𝑦′′(0) (evaluated at 𝑥 = 0)

Simplifying these expressions, we get:

4𝑦′(0) = 1 + 5𝑦′(0)

4𝑦′′(0) = 20𝑦′(0)2 + 5𝑦′′(0) + 20𝑦′′(0)

Solving the first equation for 𝑦′(0), we get:

𝑦′(0) = 1/4

Substituting this into the second equation and simplifying, we get:

4𝑦′′(0) = 20(1/4)2 + 5𝑦′′(0) + 20𝑦′′(0)

4𝑦′′(0) = 2.5 + 25𝑦′′(0)

Solving for 𝑦′′(0), we get:

𝑦′′(0) = 0.1

Therefore, 𝑦′′(0) = 0.1.