The equation is \(\displaystyle{\left({x}-{1}\right)}{y}{''}-{\left({x}+{1}\right)}{y}'+{2}{y}={0}\), and it is

London Douglas

London Douglas

Answered question

2022-04-01

The equation is (x1)y(x+1)y+2y=0, and it is given a solution type, which is polynomial. The question is to find the general solution. I looked up the general solution and it is y=C1(x2+1)+C2ex, but i don't know how to solve the question with the given data.

Answer & Explanation

ineditablesdmx0

ineditablesdmx0

Beginner2022-04-02Added 9 answers

Step 1
Considering a polynomial solution such as y1=a0+a1x+a2x2 after substitution into the DE we obtain
2a0a12a2+a1x0
so following with a1=0,2a02a2=0 we arrive at
y1=a2(x2+1)
Step 2
Assuming now that the other independent solution has the structure y2=a2(x)(x2+1) after substitution we arrive at
a2(x)=(1+2x14xx2+1)a2(x)
so now we can solve for a2(x) and then obtain y2.

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