Question

Solucionar a la siguiente ecuación de Cauchy Euler
3/2 x^2 y" +15/2 xy' + 6y = 0



Solucionar la siguiente Ecuaciones diferenciales de primer orden empleando el método de Homogéneas
5/3y" +5/2y' -5/3y = 14x^2 -4x -11

Answer 1

(1)

$y=x^m\to y'=mx^{m-1}\to y''=m(m-1)x^{m-2}\\ \\\dfrac{3}{2}x^2[m(m-1)x^{m-2}]+\dfrac{15}{2}x[mx^{m-1}]+6x^m=0\\\\m(m-1)x^{m}+5mx^m+4x^m=0\\\left[m(m-1)+5m+4\right]x^m=0\\(m^2+4m+4)x^m=0\\(m+2)^2x^m=0\\ \\\boxed{m=-2}\\ \\ \\\boxed{y=c_1x^{-2}+c_2x^{-2}\ln x}$

(2)

$\dfrac{5}{3}y'' +\dfrac{5}{2}y' -\dfrac{5}{3}y = 0\\ \\ \\2y''+ 3y'-2y=0\to \text{ecuaci\'on asociada}\to 2r^2+3r-2=0\to (2r-1)(r+2)=0\\\\r_1=\dfrac{1}{2},r_2=-2\to y_h=c_1e^{x/2}+c_2e^{-2x}\\ \\\text{Sea }y=a_2x^2+a_1x+a_0\\y'=2a_2x+a_1\\y''=2a_2\\ \\\text{Entonces tenemos}\\ \\\dfrac{5}{3}(2a_2)+\dfrac{5}{2}(2a_2x+a_1)-\dfrac{5}{3}(a_2x^2+a_1x+a_0)=14x^2 -4x -11$

$-\dfrac{5a_2}{3}x^2+(5a_2-\dfrac{5a_1}{3})x+(\dfrac{10a_2}{3}+\dfrac{5a_1}{2}-\dfrac{5a_0}{3})=14x^2 -4x -11\\ \\ \\a_2=-\dfrac{42}{5}~,~a_1=-\dfrac{114}{5}~,~a_0=-\dfrac{222}{5}\\ \\ \\y_p=-\dfrac{42}{5}x^2-\dfrac{114}{5}x-\dfrac{222}{5}\\ \\ \\y=y_h+y_p\\ \\\boxed{y=c_1e^{x/2}+c_2e^{-2x}-\dfrac{42}{5}x^2-\dfrac{114}{5}x-\dfrac{222}{5}}$