a) The given equation can be written as
(D2−1)y=1+ex2
The auxiliary equation is
m2−1=0m=±1
The complimentary function is
C.F=c1ex+c2e−x=c1f1+c2f2, where f1=ex,f2=e−x
The particular integral is = Pf1+Qf2, where
P=−∫W(f1,f2)f2g(x)dx,Q=∫W(f1,f2)f1g(x)dx
where, g(x)=1+ex2
W(f1,f2)=f1f2′−f1′f2=ex(−e−x)−exe−x=−2
P=−∫−2(1+ex)2e−xdx=∫1+exe−xdx=∫1+e−xe−2xdx
Put u=1+e−x , then du=−e−xdx
dx=−y−1dy .Therefore,
P=∫y(y−1)(y−1)2 (−dy)=−∫yy−1dy=∫(y1−1)dy=lny−yP=ln(1+e−x)−1−e−x
Pf1=ex(ln(1+e−x)−e−x−1)=ex(ln(1+e−x)−1)−1
Q=∫−2(1+ex)2exdx=−∫1+exexdxQ=−ln(1+ex)Qf2=−e−xln(1+ex)Hence,P.I=ex(ln(1+e−x)−1)−e−xln(1+ex)−1.
The complete solution is
y=C.F.+P.I=ex(c1+ln(1+e−x)−1)+e−x(c2−ln(1+ex))−1
b) The given equation can be written as, (D2+D−2)y=−6sin2x−18cos2x,y(0)=2,y′(0)=2.
The auxiliary equation is,
m2+m−2=0(m+2)(m−1)=0m=−2,1C.F=c1e−2x+c2ex
P.I=D2+D−21(−6sin2x−18cos2x)=−6D2+D−21(sin2x+3cos2x)=−6(D2+D−21(sin2x)+D2+D−21(3cos2x))=−6(−4+D−21(sin2x)+3−4+D−21(cos2x)) (Replace D2 by−4)=−6(−6+D1(sin2x)+3−6+D1(cos2x))=6(6−D1(sin2x)+36−D1(cos2x))=6(36−D26+D(sin2x)+336−D26+D(cos2x))=203(6+D)(sin2x+3cos2x) (Replace D2 by−4)=203(6sin2x+18cos2x)+D(sin2x+3cos2x)=203(6sin2x+18cos2x+2cos2x−6sin2x)P.I=3cos2x
The complete solution is
y(x)=c1e−2x+c2ex+3cos2x
Given the initial conditions, y(0)=2,y′(0)=2. We get,
2=y(0)=c1+c2+3c1+c2=−1Now,y′(x)=−2c1e−2x+c2ex−6sin2x2=y′(0)=−2c1+c2−2c1+c2=2
Solving for c1,c2 we get, c1=−1,c2=0.
Thus, the complete solution becomes,
y(x)=3cos2x−e−2x
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