Question

Answer the questions and provide complete solution

1.5) Find the largest constant r so that phi(x) = e^rz is a solution to third derivative of y minus 2 times second derivative of y plus 2y = 0.

1.6) Find the smallest constant rso that phi(x) = x^r is a solution to (3)(x^2)(second derivative of y) + (11)(x)(first derivative of y) - 3y = 0, x > 0.

Answer 1

1.5) Differential equation is $y''' - 2y'' + 2y=0$ and $\phi(x)= e^{rx}$ is solution of this differential equation, hence $\phi(x)$ holds the differential equation.

Now,

$y = \phi(x) = e^{rx}, y' = r e^{rx}, y'' = r^2 e^{rx} + e^{rx}, \\ y''' = r^3e^{rx} + 2re^{rx} +re^{rx} = r^3e^{rx} + 3re^{rx}$.

Hence by putting in differential equation, we ger

$e^{rx} (r^3+3r - 2r^2 -2 +2) = 0$

Since, $e^{rx} \neq 0, \implies r^3-2r^2 + 3r =0$

$\implies r(r^2-2r+3) =0 \implies r=0, r = \frac{2\pm \sqrt{4-12}}{2} = 1\pm\sqrt{2} i$ .

Two roots are imaginary, so real largest constant is 0.

1.6) Differential equation is $3x^2y''+11xy'-3y =0$ .

$y = x^n$ is solution when it holds the differential equation.

Now, $y = x^n, y' = nx^{n-1}, y'' = n(n-1) x^{n-2}$

By putting, we get

$\implies (3n(n-1)+11n-3)x^n=0 \implies 3n^2 +8n-3=0 \\ So, \ 3n^2+9n-n-3=0 \implies (3n-1)(n+3)=0 \\ \implies n = \frac{1}{3}, -3.$

Smallest n is -3.