User:Egm6321.f11.team2.kim/PR6

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Given
$$\displaystyle (x+1)y''-(2x+3)y' + 2y = 0 $$ $$\displaystyle y = e^{rx} $$

Find
Find $$\displaystyle u_1(x), u_2(x) $$ of eq 6.5.1 using the trial solution eq 6.5.2

Solution
$$\displaystyle y = e^{rx} y' = re^{rx} y'' = r^2e^{rx} $$ plugging eq 6.5.3 into eq 6.5.1 $$\displaystyle (x+1)r^2e^{rx}-(2x+3)re^{rx}+2e^{rx} = 0 $$ $$\displaystyle (x+1)r^2 - (2x + 3)r + 2 = 0 $$ $$\displaystyle r = \frac {(2x+3) \pm \sqrt{(2x+3)^2-8(x+1)}}{2(x+1)} $$ $$\displaystyle r = \frac{(2x+3) \pm (2x+1)}{2(x+1)} $$ $$\displaystyle r_1 = 2, \,\,\,\,r_2 = \frac{(2x+3)-2x - 1 }{2(x+1)} $$ Now we can find u(x), that is $$\displaystyle u_1(x)= e^{2x} $$ From lecture note 34-5, $$\displaystyle u_2(x)= u_1(x) \cdot \int \frac{1}{u_1^2(x)}exp[-\int a_1(x)dx]dx $$ $$\displaystyle a_1(x) = \frac {-(2x+3)}{x+1}$$ $$\displaystyle u_2(x) = e^{2x} \cdot \int{ e^{-4x}\cdot e^{\int \frac{2x+3}{x+1}dx}dx} $$ $$\displaystyle u_2(x) = e^{2x}\cdot \int{ e^{-4x}\cdot e^{(2x + log (x+1))}dx}$$ $$\displaystyle u_2(x) = e^{2x}\cdot \int{ e^{-4x}\cdot e^{2x} \cdot (x+1)dx}$$ $$\displaystyle u_2(x) = e^{2x}\cdot \int{ e^{-2x}\cdot (x+1)dx}$$ $$\displaystyle u_2(x) = e^{2x}\cdot e^{-2x}\cdot \frac {-(2x+3)}{4}$$ $$\displaystyle u_2(x) = \frac {-(2x+3)}{4} $$ Here -1/4 is constant. Thus, homogeneous solution is that $$\displaystyle y(x) = c_1\cdot e^{2x} + c_2\cdot (2x+3) $$

=R*6.5 - =