Consider the following ODE:

\begin{equation} x^2y''+6xy'+6y=\sqrt x \end{equation}

For the following question I believe that I am supposed to use "reduction of order".

(a) Verify that $y_1=x^{-2}$ is a solution to the homogeneous equation,

\begin{equation} x^2y''+6xy'+6y=0 \end{equation}

(b) By setting $y=u(x)y_1$, rewrite the ODE as a differential equation in $u(x)$. By setting $v(x)=u'(x)$, show that the new equation for $v(x)$ is first order. Thus, we have reduced the order of the ODE from two to one.

(c) Solve for $v(x)$, and hence find the general solution for the first ODE.

So, when I initially began to work I immediately got stuck on part (a), as I am not sure where to start. I think I would be able to solve part (c), but I am confused about (a) and (b). Any help would be much appreciated. Thanks in advance.

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    $\begingroup$ What is $y_1$? It should be the same as $y$. $\endgroup$ – tmaj Aug 29 '19 at 5:07
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    $\begingroup$ The homogeneous basis solutions are $x^{-2},x^{-3}$. A particular solution is obtained as $Cx^{1/2}$ with $C(-\frac14+3+6)=1\implies C=\frac4{35}$, $y(x)=Ax^{-3}+Bx^{-2}+\frac4{35}x^{1/2}$. See Euler-Cauchy equation. $\endgroup$ – Lutz Lehmann Aug 29 '19 at 9:15

$$x^2y''+6xy'+6y=\sqrt x\tag1$$ Homogeneous equation is $$x^2y''+6xy'+6y=0\tag2$$

  • Let $~y_1=x^{-2}~$, then $~y'_1=-2x^{-3}~$ and $~y''_1=6x^{-4}~$

Now $$x^2y_1''+6xy_1'+6y_1=x^2(6x^{-4})+6x(-2x^{-3})+6(x^{-2})=6x^{-2}-12x^{-2}+6x^{-2}=0$$ So $~y_1=x^{-2}~$ is solution of $(2)$.

  • Let $~y=u(x)y_1=x^{-2}~u(x)~$, then $~y'=-2x^{-3}u+x^{-2}~u'~$ and $~y''_1=6x^{-4}u-4x^{-3}~u'+x^{-2}~u''~$

Putting these values in equation $(1)$, $$x^2~(6x^{-4}u-4x^{-3}~u'+x^{-2}~u'')+6x~(-2x^{-3}u+x^{-2}~u')+6~x^{-2}~u=\sqrt x$$ $$\implies (6x^{-2}u-4x^{-1}~u'+u'')+(-12x^{-2}u+6x^{-1}~u')+6~x^{-2}~u=\sqrt x$$ $$\implies u''+2x^{-1}u'=\sqrt x\tag3$$ Putting $~v(x)=u'(x)~$ in $(3)$, $$v'+2x^{-1}v=\sqrt x\tag4$$which is a first order differential equation. Thus, we have reduced the order of the ODE from two to one.

  • Integrating factor (I.F.) $~=e^{\int 2x^{-1}dx}=x^2~$

Multiplying both side of $(4)$ by I.F. and then integrating we have $$x^2v=\frac{2}{7}x^{\frac{7}{2}}+c$$ $$\implies v=\frac{2}{7}x^{\frac{3}{2}}+cx^{-2}$$where $~c~$ is constant.

So $$u'=\frac{2}{7}x^{\frac{3}{2}}+cx^{-2}$$ $$\implies u=\frac{2}{7}\cdot \frac{2}{5}x^{\frac{5}{2}}-cx^{-1}+d$$ where $~d~$ is a constant.

Hence the solution of the equation $(1)$ is $$y=x^{-2}u=\frac{4}{35}x^{\frac{1}{2}}+a~x^{-3}+b~x^{-2}$$where $~a(=-c),~b(=d)~$ are arbitrary constants.


$$y_1=x^{-2}$$ $$y'_1=-2x^{-3}$$ $$y''_1=6x^{-4}$$ $$x^2y''_1+6xy'_1+6y_1=x^2(6x^{-4})+6x(-2x^{-3})+(6x^{-2})$$ After simplification : $$x^2y''_1+6xy'_1+6y_1=6x^{-2}-12x^{-2}+6x^{-2}$$ $$x^2y''_1+6xy'_1+6y_1=0$$ Thus $y_1(x)$ is a solution of the homogeneous ODE : $$x^2y''+6xy'+6y=0$$


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