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:
$\displaystyle \frac{dy}{(y-3)^2} = sin(3x) \:dx$
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
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
$\displaystyle -\frac{1}{0-3} = \frac{1}{3} = -frac{1}{3}cos(3) + C$
$\displaystyle C = \frac{1}{3}(1+cos(3))$