Solve the intial value problem using integrals:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]dy/dx[/tex] = (cube root of x) , y(1) = 2

The solution:

if f(x) = F(x), and y(x) is equal to F(x), then,

y(x) = ([tex]x sub 0[/tex], [tex]y sub 0[/tex])

Here's what I did, and failed:

y(1) = 2. This simply means (and I understand this) that if x is 1, then the function y(x) is 2 from the input of x equal to 1.

So, in order to find the slope of the curve at the cube root of x, when y(1) = 2. We must integrate.

this is the problem I am having.

I get:

y = [tex]integral[/tex] cube root of x [tex]dx[/tex] + C

[tex]dy/dx[/tex] = cube root of 1 (substitued the intial value of x) + C

At the coordinates (1,2)

What am I doing wrong here?

I dont' know how to integral the cube root of 3. Amd what is C? 2?

Thank you.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Initial Value systems.

Loading...

Similar Threads for Initial Value systems |
---|

I Derivative of infinitesimal value |

A Integral of Dirac function from 0 to a... value |

I Values of Lagrange multipliers when adding new constraints |

I Asymptotic expansion integral initial step |

I Finding value of parameters to fit some data |

**Physics Forums | Science Articles, Homework Help, Discussion**