# 2nd Order Ordinary Differential Equation

Consider the following ODE:

$$$$x^2y''+6xy'+6y=\sqrt x$$$$

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,

$$$$x^2y''+6xy'+6y=0$$$$

(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.

• What is $y_1$? It should be the same as $y$. – tmaj Aug 29 at 5:07
• 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. – LutzL Aug 29 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 : $$x^2y''+6xy'+6y=0$$