Answer to Question #199509 in Differential Equations for rajkumar

Question

do the functions y_{1 (t)=} √ t and y ₂ ( t ) =1\ t form a fundamental sets of solutions of the equation 2t ² y" + 3t y' - y=0, on the interval 0 <t< ∞ ? justify your answer.

Answer 1

Let $y=t^\alpha$. Then

$$y'=\alpha t^{\alpha-1}$$

$$y''=\alpha(\alpha-1) t^{\alpha-2}$$

Substitute

$$2t^2\alpha(\alpha-1) t^{\alpha-2}+3t\alpha t^{\alpha-1}-t^\alpha=0$$

$$(2\alpha^2-2\alpha+3\alpha-1)t^\alpha=0$$

$$(2\alpha^2+\alpha-1)t^\alpha=0$$

This is true for $0<t<\infin,$ if

$$2\alpha^2+\alpha-1=0$$

$$2\alpha^2+2\alpha-\alpha-1=0$$

$$2\alpha(\alpha +1)-(\alpha +1)=0$$

$$2(\alpha +1)(\alpha-\dfrac{1}{2})=0$$

$$\alpha_1=\dfrac{1}{2}, \alpha_2=-1$$

$$y_1=t^{1/2}=\sqrt{t}$$

$$y_2=t^{-1}=\dfrac{1}{t}$$

Hence $y_1(t)=\sqrt{t}$ and $y_2(t)=\dfrac{1}{t}$ form a fundamental sets of solutions of the equation 2t ² y" + 3t y' - y=0, on the interval 0 <t< ∞ .