Solving 3rd Order ODE by Reduction of Order x^3y''' -3x^2y'' +(6-x^2)xy'

Taliyah Spencer

Taliyah Spencer

Answered question

2022-04-19

Solving 3rd Order ODE by Reduction of Order
x3y3x2y+(6x2)xy(6x2)y=0

Answer & Explanation

Elliot Roy

Elliot Roy

Beginner2022-04-20Added 14 answers

Step 1
Searching for the other homogeneous solutions by assuming it is the product of some function times another known solution is called variation of parameters. Formally we are plugging in the guess y=vy1=xv into the equation D[y]=0
{y=xvy=v+xvy=2v+xvy=3v+xv
xD[v]+(3x3v6x2v+(6x2)xv)=0
(x4v3x3v+(6x2)x2v(6x2)xv)+(3x3v6x2v+(6x2)xv)
=x4(vv)=0
In other words v is the three homogeneous solutions to the above differential equation
v=C1ex+C2ex+C3
Your professor was justified in calling this a reduction of order because vv=0 is secretly a second order differential equation, with the constant solution already represented by the first homogeneous solution y1=x. This gives us our final solutions
y=C1x+C2xex+C3xex
Savanah Wilkerson

Savanah Wilkerson

Beginner2022-04-21Added 13 answers

Step 1
It will be easier if you use the formula for the second solution:
y2=v(x), where x is your first solution.
Then
fy2=vx
y2=vx+v
y2 =v x+2v
y2 =v x+3v 
Insert for every respective term of y in the original ode and get:
x3(v x+3v )3x2(v x+2v)+(6x2)x(vx+v)(6x2)vx=0
which gives
x4v +3x3v 3x3v 6x2v+(6xvx3v)x=0
x4v +3x3v 3x2v 6x2v+6x2vx4v=0
Sum up each related term and cancel out 6x2v+6x2v and 3x3v 3x3v  and obtain
x4v x4v=0
Divide by x4 and get the simplified form:
v v=0
Use reduction of order by letting u =v  and get:
u u=0
This has the simple solution
λ=±41u(x)=e2x+e2x. Replace u(x) back to v by integrating it once and get:
v(x)=u(x) dx =e2x+e2x dx =12e2x12e2x+C
At last, go back to the form of
y2=v(x)xy2(x)=x2e2xx2e2x+Cx
Nicely put:
y2(x)=x2(e2xe2x)+Cx

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