k2+3k+2=0
k1=−2
k2=−1
y(x)=C1ek1x+C2ek2x
y(x)=C1e−2x+C2e−x
e−xdxdC2(x)+e−2xdxdC1(x)=0
dxd(e−x)dxdC2(x)+dxd(e−2x)dxdC1(x)=sin(ex)
e−xdxdC2(x)+e−2xdxdC1(x)=0
−e−xdxdC2(x)−2e−2xdxdC1(x)=sin(ex)
dxdC1(x)=−e2xsin(ex)
dxdC2(x)=exsin(ex)
C1(x)=−∫e2xsin(ex)dx=excos(ex)−sin(ex)+C3
C2(x)=∫exsin(ex)dx=C4−cos(ex)
Answer:
y(x)=C3e−2x+C4e−x−e−2xsin(ex)
Comments