You are here:

- Home
- Science
- Mathematics
- Differential Equations
- Scott A Wilson

I am capable of solving most ordinary differential equations and a few not so ordinary, but those are few and far between.

I have taken Differential Equations at OSU. Occasionally I have assisted people with questions on the subject.

I wrote a thesis on the study of shock waves and rarefaction fans. That occurs in nature when mathematics fails to apply. See, mathematics says that when shock waves appear, there should be two solutions whereas nature knows there is only one way to be in that area. When rarefaction fans occur, the mathematics says there should be no solution, but despite this, nature still moves ahead with its own plan.

I received an MS degree at OSU in Mathemematics and a BS degree at OSU in Mathematical Science.

I graduated from OSU with a Bachelor of Science degree in mathematics with honors. Two years later, I graduated from OSU with a Master of Science degree in mathematics with honors. Between these two degrees, I have more hours in graduate courses that required to get them.

I answered many questions at OSU, down in south seattle at a college, at a church in Corvallis, OR, as Safeway in Washington, and many other areas. I have answered over 8,500 right here, but only a little over 190 of them have been in differential equations.

I have heard that it seems like fun to differentiate on contact. Perhaps it makes people oscillate wildly.

I know how to solve many homogeneous equations, I have learned how to solve 1st to 5th order differential equations, and strive to learn more about methods for solving other types of diffEQ's.

It doesn't take much to make solving differential equations quite complicated, even 1st order ones.

Some write the derivative of y as y' while other write it as dy/dx. That may not be that controlversial, but at least this block has been filled.

User | Date | K | C | P | Comments |
---|---|---|---|---|---|

fify | 05/19/16 | 10 | 10 | 10 | |

abhinav | 05/04/16 | 10 | 10 | 10 | very thanks |

John | 07/21/15 | 10 | 10 | 10 | Thank you so much, Scott. |

DEEPika | 06/26/15 | 10 | 10 | 10 | THANKS A LOT SIR! FOR SUCH A ..... |

O'Hara | 02/27/15 | 10 | 10 | 10 | Thankyou very much... God bless! |

We need to determine an equation for the total cost. The length across the river is a hypotenuse of a triangle with length x and 2. It then can be said to have length √(x²+2²). For this, the

Multiplying out the equation gives [I]dt - dx = xrt + [xr]dt. Dividing by dt gives I - dx/dt = xrt / dt + xr. Integrating with respect to t gives It - x = xrt + xrt + C. Combining terms gives

Multiplication is multiplication, as in 2*4 = 8, 13*4 + 52, 7*9 = 63, etc. When the same numbers are used as exponents, as in A^B, it means to multiply A by itself B times. To see the last response

Sorry for the delay. We take the coefficients, negate the Z row, and look for the greatest negative in the Z row. We also add a slack variable for each of the equation. The start of it is: -15 -25

Take x= 0 as where the stop light is at. This means that his current position is at -25. Since the equation has a -a/2, and a=4.5, the equation for distance is -25 + 13t with a -2.25t² when the time

Differential Equations

Answers by Expert: