Chapter 5A general solution to the corresponding homogeneous equation is given in (3) on page 243 ofthe text:yh(t)=c1e−t/2+c2e−t/6.A particular solution has the formyp(t)≡C,whichresults3(C)0+2(C)0+14C=1.2⇒C=4.8.Therefore,yp(t)≡4.8, and a general solution to the system isy(tyh(t)+yp(tc1e−t/2+c2e−t/6+4.8,x(t)=3y0(ty(t−c12e−t/2+c22e−t/6.8.We Fnd constantsc1andc2from the initial conditions,x(0) = 0.1andy(0) = 0.3 . Substitu-tion yields the system−c12+c22.8=0.1,c1+c2..3.Solving, we obtainc1=49/20,c2=−139/20, and sox(t−4940e−t/2−13940e−t/6.8,y(t4920e−t/2−
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.