Solve&nbsp;x3{y}&nbsp;‴+x{y}′−y=xln⁡(x)using shift&nbsp;x=ez&nbsp;and differential operator&nbsp;Dz=d&nbsp;dz&nbsp;.What does&nbsp;Dz=d&nbsp;dz&nbsp;&nbsp;mean?(ez)3{y}&nbsp;'''+ez{y}'−y=ezln⁡(ez)(e3z){y}'''+ez{y}'−y=ez{z}y=zre3r(r2−r−2)zr−3+ezrzr−1−zr=0Continue ?

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Solve&amp;nbsp;x3{y}&amp;nbsp;‴+x{y}′−y=xln⁡(x)using shift&amp;nbsp;x=ez&amp;nbsp;and differential operator&amp;nbsp;Dz=d&amp;nbsp;dz&amp;nbsp;.What does&amp;nbsp;Dz=d&amp;nbsp;dz&amp;nbsp;&amp;nbsp;mean?(ez)3{y}&amp;nbsp;&#039;&#039;&#039;+ez{y}&#039;−y=ezln⁡(ez)(e3z){y}&#039;&#039;&#039;+ez{y}&#039;−y=ez{z}y=zre3r(r2−r−2)zr−3+ezrzr−1−zr=0Continue ?

Karsyn Wu · Accepted Answer

Explanation: Step 1Let&amp;nbsp;Y(z)=y(x)=y(ez). Then Chain Rule and Product Rule givedYdz&amp;nbsp;=dydxdxdz=dydxez=xdydxd2Ydz2&amp;nbsp;=ddz(dydxez)=dydxez+d2ydx2ezez=dydxez+d2ydx2e2z=dYdz+x2d2ydx2d3Ydz3&amp;nbsp;=ddz(dYdz+d2ydx2e2z)=d2Ydz2+2d2ydx2e2z+d3ydx3e3z&amp;nbsp;&amp;nbsp;=d2Ydz2+2x2d2ydx2+x3d3ydx3=d2Ydz2+2(d2Ydz2−dYdz)+x3d3ydx3&amp;nbsp;&amp;nbsp;=3d2Ydz2−2dYdz+x3d3ydx3.Step 2Thus, the original DE transforms to                                        x            3                                y            ‴                    +          x                      y            ′                    −          y                                    =          x          ln          ⁡          x                                                  Y            ‴                    −          3                      Y            ″                    +          2                      Y            ′                    +                      Y            ′                    −          Y                                    =                      e            z                    z                                                  Y            ‴                    −          3                      Y            ″                    +          3                      Y            ′                    −          Y                                    =          z                      e            z                    .

Endstufe5qa2 · Accepted Answer

Step 1Treating the homogeneous equation as Euler-Cauchy equation, that is, trying y=xr, gives the equation0=r(r−1)(r−2)+r−1=r3−3r+3r−1=(r−1)3,so that the basis solutions are y=x,x=ln⁡(x)x,y=ln⁡(x)2x.Step 2The right side is in resonance of degree 2, thus the particular solution has the formyp(x)=ln⁡(x)3(A+Bln⁡(x))x.

Solve \(\displaystyle{x}^{{{3}}}{\left\lbrace{y}\right\rbrace}{'''}+{x}{\left\lbrace{y}\right\rbrace}'-{y}={x}{\ln{{\left({x}\right)}}}\) using shift \(\displaystyle{x}={e}^{{{z}}}\) and differential

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