Characteristic Equations And Ivp (Solutions)

CHARACTERISTIC EQUATIONS AND IVPFind the general solution to the equation (using the characteristic polynomials).2y'''+y''+2y'=0→2r3+r2+2r=0since f(r) do not have a constant term and the term with the lowest power is zr, thenf(r)r=2r2+r+2=0Then, r=0. To find the remaining roots, we can apply the quadratic formular=-b±b2-4ac2a=-1±12-4(2)(2)2(2)=-1±15i4=-14±i154with roots r=0,-14+i154,-14-i154Thus, the general solution is given by: yG=c1er1x+c2er2x+⋯+cnernxyG=c1+c2e-1/4xcos⁡(15x/4)+c3e-1/4xsin⁡(15x/4)Using Laplace transforms, solve the IVPy''-3y'+2y=20sin2t, y(0)=0,y'(0)=0Ly''-3y'+2y=L{20sin⁡(2t)}→as2+a2,s>0s2Y(s)-sy(0)-y'(0)-3[sY(s)-y(0)]+2Y(s)=202s2+22Substitute the terms,s2Y(s)-0-0-3[sY(s)-0]+2Y(s)=40s2+4Y(s)s2-3s+2=40s2+4y(s)=40s2+4s2-3s+240s2+4s2-3s+2=40s2+4(s-1)(s-2)=As+Bs2+4+C(s-1)+D(s-2)40=As+B(s-1)(s-2)+Cs2+4(s-2)+Ds2+4(s-1)40=As3-3As2+2As+Bs2-3Bs+2B+Cs3-2Cs2+4Cs-8c+Ds3-Ds2+4Ds-4DThus, Coefficient of s3:0=A+C+Ds2:0 =-3A+B-2C-Ds:0 =2A-3B+4C+4D const: 40 =2B-8C-4DUse the coefficient matrix and vector for this system ofx=1011-31-2-12-34402-8-4 y=00040A=xy,1|x|=001101-2-10-344402-8-41011-31-2-12-34402-8-4=12040=3B=xy,2|x|=1011-30-2-120440408-440=-8040=-2C=xy,3|x|=1001-310-12-3040240-440=-32040=-8D=xy,4x=1010-31-202-34002-84040=-20040=5We get A=3,B=-2,C=-8,D=5. Substituting into the equation,40s2+4(s-1)(s-2)=3s-2s2+4-8(s-1)+5(s-2)We can now apply the inverse transform,L-1{y(s)} =L-13ss2+4ss2+a2-2s2+4as2+a2-2s-1eat+5s-2eat }y(<w:highlight