Solve dx2d2y(x)−y(x)=3exx2The general solution will be the sum of the complementary solution and particular solution.Find the complementary solution by solving dx2d2y(x)−y(x)=0:Assume a solution will be proportional to eλx for some constant λ. Substitute y(x)=eλx into the differential equation:dx2d2(eλx)−eλx=0 Substitute dx2d2(eλx)=λ2eλx:x2exx−exx=0 Factor out eλx:(λ2−1)eλx=0
Since eλx=0 for any finite λ, the zeros must come from the polynomial:λ2−1=0 Factor:(λ−1)(λ+1)=0Solve for λ:λ=−1 or λ=1The root λ=−1 gives y1(x)=c1e−x as a solution, where c1 is an arbitrary constant.The root λ=1givesy2(x)=c2ex as a solution, where c2 is an arbitrary constant. The general solution is the sum of the above solutions:y(x)=y1(x)+y2(x)=c1e−x+c2exDetermine the particular solution to dx2d2y(x)−y(x)=3exx2 by the method of undetermined coefficients:The particular solution to dx2d2y(x)−y(x)=3exx2 is of the form:yp(x)=x(a1ex+a2exx+a3exx2), where a1ex+a2exx+a3exx2was multiplied by x to account for ex in the complementary solution.
Solve for the unknown constants a1,a2,anda3: Compute dx2d2yp(x):dx2d2yp(x)=dx2d2(a1exx+a2exx2+a3exx3)=a1(2ex+exx)+a2(2ex+exx2+4exx)+a3(exx3+6exx2+6exx)Substitute the particular solutionyp(x) into the differential equation:dx2d2yp(x)−yp(x)=3exx2a1(2ex+exx)+a2(2ex+exx2+4exx)+a3(exx3+6exx2+6exx)−(a1exx+a2exx2+a3exx3)=3exx2 Simplify:(2a1+2a2)ex+(4a2+6a3)exx+6a3exx2=3exx2Equate the coefficients of ex on both sides of the equation:2a1+2a2=0 Equate the coefficients of exx on both sides of the equation:4a2+6a3=0
Equate the coefficients of exx2 on both sides of the equation:6a3=3 Solve the system:a1=43a2=−43a3=21Substitute a1,a2, and a3 into yp(x)=exxa1+exx2a2+exx3a3:yp(x)=2exx3−43exx2+43exx The general solution is:y(x)=yc(x)+yp(x)=2exx3−43exx2+43exx+c1e−x+c2ex
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