Using the method of undetermined coefficients, find the general solution of the following differential equation y″−3y′+2y=x2.

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habbocowji · Accepted Answer

We find the general solution of eq (1) of the method of undesermened coefficient.Now the aurilary eq for the homogenous eqy3y′+2y=0 ism2−3m+2=0(m−2)(m−1)=0m=1,2So the function y=c1e+c2e2xHere we take yp=Ax2+Bx+c,yp=2a+byp2aPutting the value of yp,yp,yp in eq (1) we get2a−3(2ax+b)+2(ax2+bx+c)=x22a−6ax−3b+2ax2+2bx+2=x22ax2+(−6a+2b)x+(2a−3b+2c)=x22a=1⇒a=12−6A+2b=0⇒−6×12+2b=0⇒−3+2b=0⇒b=322a−3b+2c=0⇒2⋅12−3⋅32+2=0⇒1−92+2c=0⇒2c⋅92−12c=12=1c=14yp=12x2+32x+14The general solution isy(x)=yc+yp=c1ex+c2e2x+12x2+32x+14

Deufemiak7 · Accepted Answer

Given differential equation isy3y′+2y=x2+x+1 (1)AE is D2−3D+2=0(D−1)(D−2)=0∴D=1,2∴CF=c1ex+c2e2xPI is of the formy=A0+A1x+A2x2y′=0+A1+2A2xy0+2A2Substituting in (1)2A2−3(A1+2A2x)+2(A0+A1x+A2x2)=x2+x+1∴2A2x2+(2A1−6A2)x+2A0−3A1+2A2=x2+x+1Comparing the coefficients2A2=1,2A1−6A2=1,2A0−3A1+2A2=1Solving, A2=12,A1=2,A0=3∴PI=3+2x+12x2∴GS=CF+PIy=c1ex+c2e2x+(3+2x+12x2)

Vasquez · Accepted Answer

Solve y″−3y′+2y=x2Homogen solution: y=Aex+Be2xParticular solution:yp=x(Ax2+Bx+C)=Ax3+Bx2+Cxy′p=3Ax2+2Bx+Cy″p=6Ax+2BPut his into the initial equartion to get A, B and C gives me:A=0,B=1/2,C=3/2This leads me to the answer:y=Aex+Be2x+x2/2+3x/2However the correct answer isy=Aex+Be2x+x2/2+3x/2+7/4Where&#039;s my miss? Where comes the last term from?

Using the method of undetermined coefficients, find the general solution

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