Question

Do the functions y1(t)=root t and y2(t)=1/t form a fundamental set of solutions of the equation 2t^2 y'' +3t y' -y=0, on the interval 0 less t less ~? justify your answer.

Answer 1

Answer on Question #74058 – Math – Differential Equations

Question

Do the functions $y_{1}(t) = \sqrt{t}$ and $y_{2}(t) = 1 / t$ form a fundamental set of solutions of the equation $2t^{2}y^{\prime \prime} + 3ty^{\prime} - y = 0$, on the interval $t\in (,\infty)$?

Justify your answer.

Solution

$$2 t ^ {2} y ^ {\prime \prime} + 3 t y ^ {\prime} - y = 0$$

This is a Cauchy-Euler equation which is solved by the substitution $y = t^r$.

Differentiating we have

$$y ^ {\prime} = r t ^ {r - 1}$$ $$y ^ {\prime \prime} = r (r - 1) t ^ {r - 2}$$

Then

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

Use the quadratic formula

$$r = \frac {- 1 \pm \sqrt {1 - 4 (2) (- 1)}}{2 (2)} = \frac {- 1 \pm 3}{4}$$

We have two distinct real roots

$$r _ {1} = \frac {- 1 + 3}{4} = \frac {1}{2}$$ $$r _ {2} = \frac {- 1 - 3}{4} = - 1$$

The fundamental set of solutions of the equation

$$y = c _ {1} (\sqrt {t}) + c _ {2} \left(\frac {1}{t}\right), t > 0$$

Yes, the functions $y_{1}(t) = \sqrt{t}$ and $y_{2}(t) = 1 / t$ form a fundamental set of solutions of the equation $2t^{2}y^{\prime \prime} + 3ty^{\prime} - y = 0$, on the interval $t\in (,\infty)$.

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