Let us solve the differential equation x2y′′−2xy′−4y=x2+2lnx.
Let us use the transformation x=et. Then yx′=yt′e−t, yx2′′=(yt2′′−yt′)e−2t.
We get the equation e2t(yt2′′−yt′)e−2t−2etyt′e−t−4y=e2t+2t, which is equivalent to yt2′′−3yt′−4y=e2t+2t.
The characteristic equation k2−3k−4=0 of the differential equation yt2′′−3yt′−4y=0 is equivalent to (k+1)(k−4)=0, and hence has the roots k1=−1, k2=4.
It follows that the general solution of the differential equation yt2′′−3yt′−4y=e2t+2t is y(t)=C1e−t+C2e4t+yp, where yp=ae2t+bt+d. Then yp′=2ae2t+b, yp′′=4ae2t.
It follows that
4ae2t−3(2ae2t+b)−4(ae2t+bt+d)=e2t+2t, and hence
−6ae2t−4bt−3b−4d=e2t+2t.
We conclude that −6a=1, −4b=2, −3b−4d=0, and thus a=−61, b=−21, d=−43b=83.
Therefore, the general solution of the differential equation yt2′′−3yt′−4y=e2t+2t is y(t)=C1e−t+C2e4t−61e2t−21t+83.
We conclude that the general solution of the differential equation x2y′′−2xy′−4y=x2+2lnx is y(x)=C1x−1+C2x4−61x2−21lnx+83.
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