dy/dx=(y-3)^2sin(3x) with the initial condition of f(1)=0. Find the particular solution. I need serious help!

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- May 21st 2009, 10:38 AMxvaliantInitial Value Problem?
dy/dx=(y-3)^2sin(3x) with the initial condition of f(1)=0. Find the particular solution. I need serious help!

- May 21st 2009, 10:45 AMe^(i*pi)
Separate the Variables:

$\displaystyle \frac{dy}{(y-3)^2} = sin(3x) \:dx$

Let $\displaystyle u = y-3$ so that $\displaystyle du = dy$

$\displaystyle \int \frac{du}{u^2} = \int sin(3x) \:dx$

When you solved that put in x=1 and y=0 to find a value for the constant of integration, C - May 21st 2009, 02:48 PMxvaliant
- May 21st 2009, 03:59 PMe^(i*pi)
Where did you get the exponential from?

$\displaystyle \int frac{du}{u^2} = -\frac{1}{u}$

As u = y-3: $\displaystyle -\frac{1}{u} = -\frac{1}{y-3}$

The RHS is a standard trig integral, pull out the 1/3 to get it as $\displaystyle -\frac{1}{3}cos(3x) + C$

Overall I get:

$\displaystyle -\frac{1}{y-3} = -\frac{1}{3}cos(3x) + C$

Since f(1)=0

$\displaystyle -\frac{1}{0-3} = \frac{1}{3} = -frac{1}{3}cos(3) + C$

$\displaystyle C = \frac{1}{3}(1+cos(3))$ - May 21st 2009, 04:30 PMxvaliant
Oh! I get it now. Thanks so much!