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

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

Answers (1)

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.
Substitute the values of \(y, y'\) and \(y''\) in \(x^2y''-5xy'+9y=0\) and obtain the differential equation in terms of v.
Now let \(w=v'\).Then the equation becomes \(x^5w'+x^4w=0\)
Solve the equation \(x^5w'+x^4w=0\)
Integrate \(v'=c_1x^{-1}\) and obtain the value of v.
\(v=\int c_1 x^{-1}\)
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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