# General solution to differential equation

• Nov 11th 2009, 10:40 PM
jsl
General solution to differential equation
Hi! I'm having trouble with this problem on my homework. Any help you can give would be wonderful!! Thanks!

Consider the differential equation:
http://webwork2.math.utah.edu/webwor...1b7c82c891.png.
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 http://webwork2.math.utah.edu/webwor...c23ba91971.png --- rename your constant if necessary.)

b) Find the particular solution of the above differential equation that satisfies the condition http://webwork2.math.utah.edu/webwor...124b08cd21.png at http://webwork2.math.utah.edu/webwor...388a5f95d1.png.
• Nov 11th 2009, 10:48 PM
Drexel28
Quote:

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

Consider the differential equation:
http://webwork2.math.utah.edu/webwor...1b7c82c891.png.
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 http://webwork2.math.utah.edu/webwor...c23ba91971.png --- rename your constant if necessary.)

b) Find the particular solution of the above differential equation that satisfies the condition http://webwork2.math.utah.edu/webwor...124b08cd21.png at http://webwork2.math.utah.edu/webwor...388a5f95d1.png.

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 $y=\frac{-4}{x^4+2x^2+C_1}$
• Nov 13th 2009, 04:32 PM
shalaylee
actually:

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