Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 6 / c

Table of contents
- Solution c

Solution

Divide by

x^{2}

, which is allowed because

x > 0

:

y^{'} + \frac{1}{x} y = 4 e^{x^{2}} .

The integrating factor for this linear ordinary differential equation is

e^{\int d x / x} = x

. Multiplying through makes the left side a product rule derivative:

(x y)^{'} = 4 x e^{x^{2}} .

Part (b) now gives

x y = 2 e^{x^{2}} + C

. Therefore

y (x) = \frac{2 e ^{x^{2}} + C}{x}, x > 0.

 Ancestors (11)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version