Reduce to first order and solve: x^2y''-5xy'+9y=0 y_1=x^3

Question
Reduce to first order and solve:
\(x^2y''-5xy'+9y=0\ \ y_1=x^3\)

Answers (1)

2020-12-25
The differential equation is \(x^2y''-5xy'+9y=0\) and one solution is \(y_1=x^3\)
Now, let the new solution be \(y=vy_1=vx^3\) where v is again a function of x.
Then,
\(y=vx^3\)
\(y'=3vx^2+x^3v'\)
\(y''=3(2vx+x^2v')+x^3v''+3x^2v'\)
\(=6vx+3x^2v'+x^3v''+3x^2v'\)
\(=6vx+6x^2v'+x^3v''\)
Substitute the values of \(y, y'\) and \(y''\) in \(x^2y''-5xy'+9y=0\) and obtain the differential equation in terms of v.
\(x^2(6vx+6x^2v'+x^3v'')-5x(3vx^2+x^3v')+9vx^3=0\)
\(6vx^3+6x^4v'+x^5v''-15x^3v-5x^4v'+9vx^3=0\)
\(x^5v''+x^4v'=0\)
Now let \(w=v'\).Then the equation becomes \(x^5w'+x^4w=0\)
Solve the equation \(x^5w'+x^4w=0\)
\(x^5\frac{dw}{dx}+x^4w=0\)
\(x^5\frac{dw}{dx}=-x^4w\)
\(\frac{dw}{w}=\frac{-1}{x}dx\)
\(\int\frac{dw}{w}=-\int\frac{1}{x}dx\)
\(\ln|w|=-\ln|x|+c\)
\(e^{\ln|w|}=e^{-\ln|x|+c}\)
\(w=c_1x^{-1}\)
\(v'=c_1x^{-1}\)
Integrate \(v'=c_1x^{-1}\) and obtain the value of v.
\(v=\int c_1 x^{-1}\)
\(=c_1\ln|x|+c_2\)
Therefore, the general solution is \(y=c_1x^3\ln x +c_2 x^3\)
The second solution is \(y_2=c_1x^3\ln|x|\)
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