Initial Value Problem?

**xvaliant** · May 21st 2009, 10:38 AM

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

**e^(i*pi)** · May 21st 2009, 10:45 AM

Originally Posted by xvaliant

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

Separate the Variables:

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

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

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

When you solved that put in x=1 and y=0 to find a value for the constant of integration, C

**xvaliant** · May 21st 2009, 02:48 PM

Originally Posted by e^(i*pi)

Separate the Variables:

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

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

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

When you solved that put in x=1 and y=0 to find a value for the constant of integration, C

Okay, so I get (y-3)^2=C(1)e^((-1/3)cos3x) Then plug in x=1 and y=0 to get C. 9=C(1)(.7168) So C(1)=12.5558? So then I plug it back into the equation to get:

(y-3)^2=12.5558e^((-1/3)cos3x)?

**e^(i*pi)** · May 21st 2009, 03:59 PM

Originally Posted by xvaliant

Okay, so I get (y-3)^2=C(1)e^((-1/3)cos3x) Then plug in x=1 and y=0 to get C. 9=C(1)(.7168) So C(1)=12.5558? So then I plug it back into the equation to get:

(y-3)^2=12.5558e^((-1/3)cos3x)?

Where did you get the exponential from?

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

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

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

Overall I get:

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

Since f(1)=0

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

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

**xvaliant** · May 21st 2009, 04:30 PM

Oh! I get it now. Thanks so much!

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