Given a second order linear homogeneous differential equation a2(x)y+a1(x)y+a0(x

Question

Given a second order linear homogeneous differential equation

a2(x)y+a1(x)y+a0(x)y=0   

we know that a fundamental set for this ODE consists of a pair linearly independent solutions y1,y2. But there are times when only one function, call it y1, is available and we would like to find a second linearly independent solution. We can find y2 using the method of reduction of order.

First, under the necessary assumption the a2(x)0 we rewrite the equation as

y+p(x)y+q(x)y=0   p(x)=a1(x)a2(x),  q(x)=a0(x)a2(x),

Then the method of reduction of order gives a second linearly independent solution as

y2(x)=Cy1u=Cy1(x)ep(x)dxy21(x)dx

where C is an arbitrary constant. We can choose the arbitrary constant to be anything we like. Once useful choice is to choose C so that all the constants in front reduce to 1. For example, if we obtain y2=C3e2x then we can choose C=1/3 so that y2=e2x.

Given the problem

9y12y+4y=0

and a solution y1=e(2x/3).

Applying the reduction of order method to this problem we obtain the following

y21(x)=

p(x)=

and ep(x)dx=

So we have

e^p(x)dx/y21(x)dx= dx=

Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at

y2(x)=Cy1u=

So the general solution to 9y12y+4y=0 can be written as

y=c1y1+c2y2=c1 +c2

Explanation / Answer

9y''-12y'+4y=0,

dividing throughout by 9,

y''-(4/3)y'+(4/9)y=0

comparing this equation with

y''+p(x)y'+q(x)y=0

we get

p(x)=-4/3

q(x)=4/9

y1(x)= e^(2x/3)

y12(x)= e(2x/3)*e(2x/3)=e(4x/3)

e-p(x).dx= e(4/3).dx=e(4x/3)

(ep(x)dx)/y21(x).dx= e(4x/3)/e(4x/3).dx= 1.dx=x

u=(ep(x)dx)/y21(x).dx=x

y2(x)= Cy1u=C.e(2x/3).x

Let C=1

y2(x)=x.e(2x/3)

y=c1y1+c2y2= c1.e(2x/3)+c2.x.e(2x/3)

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