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

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

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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