Solve the Euler equation $x^2y''-2y=9x^{0.5}$

Question

$$x^2y''-2y=9x^{0.5}$$ for $g(x)=9x^{0.5}$ I used $x=e^t$ and $q(t)=9e^{0.5t}$ instead. Now I "guessed" $$y_p=Ae^{0.5t}t^s$$ which by x is the solution $$y_p=Ax^{0.5}(\ln(x))^s$$ while $s$ is supposed to be the algebraic mult. of $0.5$ of the characteristic polynomial. While the characteristic polynomial of this Euler equation should be (the homogeneous equation) $$r(r-1)-2=0$$ $$r^2-r-2=0$$ $$(r-2)(r+1)=0$$ $$r=2 , r=-1$$ non of these eigenvalues equals $ 0.5$. What do I understand wrong? Can I say that the algebraic mult. of $0.5$ of the characteristic polynomial is zero. And $$y_p=Ae^{0.5t}t^0$$ , $$y_p=Ae^{0.5t}$$ , $$y_p=Ax^{0.5}$$ ? If yes, now I can put $y_p$ in the equation above and calculate A, then the general solution would be $$y_c=C_1x^2+C_2x^{-1}+Ax^{0.5}$$ . Am I working right?

Answer 1

$$x^2y''-2y=9x^{0.5}$$ becomes with $x=e^t$ $$y''-y'-2y=9e^{t/2}$$ Your guess is $Ae^{t/2}$ $$ \implies \frac A 4 e^{t/2}-\frac A2e^{t/2}-2Ae^{t/2}=9e^{t/2}$$ $$\implies A=-4$$ And the complete solution is $$y(t)=c_1e^{-t}+c_2e^{2t}-4e^{t/2}$$ $$y(x)=c_1x^{-1}+c_2x^{2}-4x^{1/2}$$

Answer 2

You could also let $y=x^m$. Then, $y'=mx^{m-1},y''=(m-1)mx^{m-2}$ and substituting into the homogeneous equation

$$x^2y''-2y=0$$

forms

$$x^m\big(m^2-m-2\big)=0$$

or

$$m^2-m-2=0 \implies (m-2)(m+1)=0$$

where $m_1=2,m_2=-1$. Therefore, the homogeneous solution is

$$y_h=c_1x^2+c_2x^{-1}$$

The guess for the particular solution is

$$y_p=Ax^{1/2}$$

therefore

$$y_p'=\frac{1}{2}Ax^{-1/2}$$ $$y_p''=-\frac{1}{4}Ax^{-3/2}$$

and substituting these into

$$x^2y_p''-2y_p=9x^{1/2}$$

and relating coefficients

$$-\frac{1}{4}Ax^{1/2}-2Ax^{1/2}=9x^{1/2} \implies A=-4$$ gives the particular solution as $y_p=-4x^{1/2}$. Therefore, the general solution is

$$y_g = y_h+y_p = c_1x^2+c_2x^{-1}-4x^{1/2}$$