y''-x2y'-2xy=0;y(0)=1&y'(0)=0 solve by reduction of order
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Expert's answer
2021-06-08T16:54:39-0400
We can re-write the equation as: dx2d2y(x)−x2dxdy(x)−2xy(x)=0, such that y(0)=1 and y′(0)=0We solve as follows: Substitute −2x=dxd(−x2):dx2d2y(x)−x2dxdy(x)+dxd(−x2)y(x)=0Apply the reverse product rule fdxdg+gdxdf=dxd(fg) to the left-hand side and simplify:dxd(dxdy(x))+dxd(−x2y(x))=0Factor: dxd(dxdy(x)−x2y(x))=0Integrate both sides with respect to x:∫dxd(dxdy(x)−x2y(x))dx=∫0dxEvaluate the integrals: dxdy(x)−x2y(x)=c1,⋯(i)( where c1 is an arbitrary constant.)
Let μ(x)=e∫−x2dx=e−x3/3.Multiply both sides by μ(x):e−x3/3dxdy(x)−(e−x3/3x2)y(x)=c1e−x3/3Substitute −e−x3/3x2=dxd(e−x3/3):e−x3/3dxdy(x)+dxd(e−x3/3)y(x)=c1e−x3/3Apply the reverse product rule fdxdg+gdxdf=dxd(fg) to the left-hand side:dxd(e−x3/3y(x))=c1e−x3/3Integrate both sides with respect to x:∫dxd(e−x3/3y(x))dx=∫c1e−x3/3dxEvaluate the integrals:e−x3/3y(x)=−32/33x3c1xΓ(31,3x3)+c2, where c2 is an arbitrary constant.
Divide both sides by μ(x)=e−x3/3:y(x)=ex3/3(−32/33x3c1xΓ(31,3x3)+c2)Simplify the arbitrary constants:The general solution to the differential equation is:y(x)=ex3/3(3x3c1xΓ(31,3x3)+c2)⋯(ii)
Evaluating the differential equation at y(0) = 1 and y'(0) =0.
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