# Math Help - Initial Value Problem?

1. ## Initial Value Problem?

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

2. 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

3. 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)?

4. 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

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

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

5. Oh! I get it now. Thanks so much!