Final answer:
To find y', first integrate y'' to get y' and apply the initial condition y'(0) = 1 to find the integration constant. This results in y' = -\(\frac{1}{x+1}\) + 2. Further integration would yield y, but it is not necessary for finding y'.
Step-by-step explanation:
The given differential equation is y'' = \(\frac{1}{{(x+1)}^2}\) with initial conditions y(0) = 2 and y'(0) = 1. To find the value of y', we need to integrate the second derivative of y twice.
First integration:
- Integrating y'' = \(\frac{1}{{(x+1)}^2}\), we get:
- y' = -\(\frac{1}{x+1}\) + C1,
- Using y'(0) = 1 to find the constant C1, we have:
- 1 = -\(\frac{1}{0+1}\) + C1
- C1 = 2.
Hence, y' = -\(\frac{1}{x+1}\) + 2.
Second integration:
- Integrating y' to find y, we get:
- y = -ln|x+1| + 2x + C2,
- Using y(0) = 2 to find the constant C2, we have:
- 2 = -ln|0+1| + 2(0) + C2
- C2 = 2.
Therefore, the final expression for y' in terms of x is:
y' = -\(\frac{1}{x+1}\) + 2.