# How to find fundamental set of solutions of complementary equation of a given differential equation

Given the differential equation $$4 x^2 y'' + (4x-8 x^2) y' + (4 x^2-4x-1) y = 4 x^{1/2} e^x (1+4 x).$$ I have found a general solution to a differential equation to be $$y_G = e^x ~ ( 2 x^{3/2} + x^{1/2} Ln(x) - c_1 x^{-1/2} + c_2 x^{1/2} ).$$

I am given that $$~y_1 = e^x x^{1/2}~$$ satisfies the complementary solution of the same differential equation. I need to find a fundamental set of solutions $$~\{y_1,y_2\}~$$ of the complementary equation of the original differential equation. How do I go about finding that?

One approach is to use two solutions by giving values to $$~c_1~$$ and $$~c_2~$$ and take the difference between these two solutions as another solution which becomes the second member of the fundamental set of equations or $$~y_2~$$. I don't have a method which consistently works using this approach.

$$~y_1~$$ and $$~y_2~$$ form a basis of a vector space. Which vector space are we talking about here? All vectors generated from $$~y_1~$$ and $$~y_2~$$?

How do I determine what $$~y_2~$$ is given $$~y_1~$$ for a given solution to a differential equation?

MM

• 4 x^2 y'' + (4x-8 x^2) y' + (4 x^2-4x-1) y = 4 * x^(1/2) * e^x * (1+4 x) Jul 16, 2019 at 12:15
• No. I used a reduction in order to find the general solution. I also need to find the fundamental set of solutions of the complementary equation. In the past, I have taken terms from the general solution that are linearly independent and used these as elements of the fundamental set. This time that does not appear to work. I checked Paul Dawkins' website as you recommended and formed y2=v*y1 and applied the reduction of order. I am supposed to do this as a byproduct of finding the general solution. It does not appear to be a byproduct! This is a long problem. Jul 16, 2019 at 19:30

$$4 x^2 y'' + (4x-8 x^2) y' + (4 x^2-4x-1) y = 4 x^{1/2} e^x (1+4 x).\tag 1$$ The form of the term on the right with $$e^x$$ as factor draw us to a change of function $$\quad y(x)=e^xu(x)$$ in order to eliminate the exponential term.

$$y'=e^x(u+u')$$

$$y''=e^x(u+2u'+u'')$$

$$4 x^2 (u+2u'+u'') + (4x-8 x^2) (u+u') + (4 x^2-4x-1) u = 4 x^{1/2} (1+4 x).$$ After simplification :

$$4 x^2u''+ 4x u' -u = 4 x^{1/2} (1+4 x). \tag 2$$ The solution of the homogeneous equation $$\quad 4 x^2u{_h}''+ 4x u{_h}' -u_h =0\quad$$ is easy to find : $$u_h=c_1x^{1/2}+c_2x^{-1/2}$$ Then one have to find a particular solution of Eq.$$(2)$$ (In fact any one particular solution).

One can use the method of variation of parameter http://mathworld.wolfram.com/VariationofParameters.html

A particular solution of $$\quad 4 x^2u''+ 4x u' -u = 4 x^{1/2}\quad$$ is for example $$\quad 2x^{3/2}$$.

A particular solution of $$\quad 4 x^2u''+ 4x u' -u = 16 x^{3/2}\quad$$ is for example $$\quad x^{1/2}\ln(x)$$.

Thus the general solution of Eq.$$(1)$$ is :

$$u=c_1x^{1/2}+c_2x^{-1/2}+2 x^{3/2}+x^{1/2}\ln(x)$$ $$y=e^x\big(c_1x^{1/2}+c_2x^{-1/2}+2 x^{3/2}+x^{1/2}\ln(x)\big)$$