x^2y''+(3x^2+4x)y'+2(x^2+3x+1)y=0 I try to solve this equation by finding \mu such

Ali Marshall 2022-05-01 Answered
x2y+(3x2+4x)y+2(x2+3x+1)y=0
I try to solve this equation by finding μ such that the equation become exact.( I know there are other ways for solving this equation).
(μy)=μy+μy=μy+μ3x2+4xx2y+μ2(x2+3x+1)x2y
μy=μ3x2+4xx2y+μ2(x2+3x+1)x2yμμ=3x2+4xx2+2(x2+3x+1)x2yy
How am I supposed to solve it?
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Answers (1)

Jonas Dickerson
Answered 2022-05-02 Author has 22 answers

Step 1
There is a method to solve equations like this one that uses a clever substitution. The idea is to use substitution to remove the y' term. Whenever you have an equation in the form of 
y +f(x)y+g(x)y=0, you substitute y=exp(f(x)2, dx )μ. Divide through by x2 to get
y +3x+4x=f(x)y+2(x2+3x+1)x2=g(x)y=0.
Then substitute in y=exp(3x+42x, dx )μ=e3x2x2μ.
Therefore,
e3x2x2μ e3x24x2μ=0.
Step 2
Divide through by e3x2x2 to get
μ 14μ=0.
Using standard methods in ODEs, we get
μ=c1ex2+c2ex2.
Therefore, y=e3x2x2μ=e3x2x2(c1ex2+c2ex2).

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