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Type: Open-Ended
Category: Initial Value Problems
Grade: College
Score: 1
Author: BacMan
Last Modified: 2 weeks ago

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1 vote

Solve where and
- Itegrate:
  $y' = int12x^2dx = 4x^3 + c_1$
  
  Using the first condition $y'(0) = 15$ :
  $4(0)^3 + c_1 = 15$
  $c_1 = 15$
  
  Therefore, the differential equation is:
  $y' = 4x^3 + 15$
  
  Integrate again:
  $y = int(4x^3 + 15)dx = x^4 + 15x + c_2$
  
  Using the second condition $y(0) = 13$ :
  $0^4 + 15*0 + c_2 = 13$
  $c_2 = 13$
  
  The answer is:
  $y = x^4 + 15x + 13$

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