Question

how to solve 4x^2y''-3y=0, y(1)=3, y'(1)=0, x^3/2, x^-1/2 ?

Answer 1

\[4x^2y''-3y=0\]\[y=Cx^n\quad,\quad y''=C n(n-1)x^{n-2}\]\[4x^2\cdot C n(n-1)x^{n-2}-3\cdot Cx^n=0\]\[C(4n(n-1)-3)x^n=0\]\[4n(n-1)-3=0\]\[n^2-n-3/4=0\]\[n_{1,2}=\frac{1\pm\sqrt{1+3}}{2}=\frac{1\pm 2}{2}\quad\Rightarrow\quad n_1=3/2\quad,\quad n_2=-1/2\]\[y=C_1x^{3/2}+C_2x^{-1/2}\quad\Rightarrow\quad y(1)=C_1+C_2=3\]\[y'=\frac{3}{2}C_1x^{1/2}-\frac{1}{2}C_2x^{-3/2}\quad\Rightarrow\quad y'(1)=\frac{3}{2}C_1-\frac{1}{2}C_2=0\]\[C_1=3/4\quad,\quad C_2=9/4\]