General solution to differential equation

jsl · #1 $Old$ November 11th, 2009, 10:40 PM

Hi! I'm having trouble with this problem on my homework. Any help you can give would be wonderful!! Thanks!

Consider the differential equation:

.
a) Find the general solution to the above differential equation. (Instruction: Write the answer in a form such that its numerator is 1 and its integration constant is

--- rename your constant if necessary.)

b) Find the particular solution of the above differential equation that satisfies the condition

at

.

Drexel28 · #2 $Old$ November 11th, 2009, 10:48 PM

Quote:

Originally Posted by jsl $View Post$

Hi! I'm having trouble with this problem on my homework. Any help you can give would be wonderful!! Thanks!

Consider the differential equation:

.
a) Find the general solution to the above differential equation. (Instruction: Write the answer in a form such that its numerator is 1 and its integration constant is

--- rename your constant if necessary.)

b) Find the particular solution of the above differential equation that satisfies the condition

at

.

I am just going to rewrite this as

$y'=y^2\left(x^3-x\right)$ if you don't mind.

Then $\frac{y'}{y^2}=x^3-x$ So

$\int\frac{y'}{y^2}dx=\int\left[x^3-x\right]dx$ . Evaluating gives

$\frac{-1}{y}=\frac{x^4}{4}-\frac{x^2}{2}+C$

Solving gives $y=\frac{-4}{x^4+2x^2+C_1}$

shalaylee · #3 $Old$ November 13th, 2009, 04:32 PM

actually:

(a) $4/(t^4-2t^2+4C)$
(b) $4/(t^4-2t^2+1)$