I have two functions, $y_1=x^2$ and $y_2=x^2ln(x)$. How can I prove that both are solutions of the linear homogeneous equation $x^3y'''-2xy'+4y=0$?
All I have right now is that by the superposition principle, the following linear combination is also a solution:
$$
y=c_1x^2+c_2x^2ln(x)
$$
But I'm not sure where to go from here, and the book I'm reading (Morris Tenenbaum/Harry Pollard) doesn't give similar examples. What would be the process for proving this type of exercises? Thanks in advance!
Best Answer
If $y_1$ is a solution of the differential equation: $$x^3y'''-2xy'+4y=0$$ Then you have $$x^3y_1'''-2xy_1'+4y_1=0$$ $$x^3.0-2x(2x)+4x^2=0$$ Which is true. So $x^2$ is a solution of the DE. Do the same for $y_2=x^2 \ln x$ $$x^3y_2'''-2xy_2'+4y_2=0$$