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If y''=(1)/(x+1)² with initial conditions y(0)=2 and y'(0)=1, what is the value of y'?

User Jillan
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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.

User Nicholas White
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