User:Egm6321.f09.team7.hua

Hello, this is Brandon Hua. I== HW4: Problem 3 ==

Problem Statement and Given:  $$ xy''+2y'+xy=0$$ $$u_1(x)=\frac{sin(x)}{x}$$ Find $$u_2(x)$$ the second homogeneous solution

Solution: $$ xy''+2y'+xy=0$$ Divide by x to show general solution of: $$ y''+a_1y'+a_0y=0$$ $$ y''+\frac{2}{x}y'+1y=0$$ So: $$ a_1=\frac{2}{x}$$ $$ a_0=1$$

$$u_2=u_1(x)\int_{}^x \frac{1}{u_1^2(t)}e^{-\int_{}^t a_1(s)ds}dt$$ $$-\int_{}^t \frac{2}{s}ds=-2ln(t)=ln{\frac{1}{t^2}}$$ $$u_2=\frac{sin(x)}{x}\int_{}^x \frac{1}{\frac{sin^2(t)}{t^2}}e^{ln{\frac{1}{t^2}}}dt$$ $$u_2=\frac{sin(x)}{x}\int_{}^x \frac{sin^2(t)}{t^2}t^2dt$$ $$u_2=\frac{sin(x)}{x}\int_{}^x sin^2(t)dt$$ $$u_2=-\frac{sin(x)}{x}cot(x)$$ recall $$cot(x)=\frac{sin(x)}{cos(x)}$$ $$u_2=-\frac{cos(x)}{x}$$ Egm6321.f09.team7.hua 21:27, 21 October 2009 (UTC)