The given system of differential equations is,
\(Dx+Dy+(D+1)z=0\)
\(Dx+y=e^{t}\)
\(Dx+y-2z=50\sin(2t)\)
Putting the value of \(Dx+y\) in the third equation, we get,
\(Dx+y-2z=50\sin(2t)\)
\(\Rightarrow e^{t}-2z=50\sin(2t)\)
\(\Rightarrow z=\frac{1}{2}e^{t}-25\sin(2t)\)
Now differentiating both sides with respect to t, we get,
\(Dz=\frac{1}{2}e^{t}-50\cos(2t)\)
\(\Rightarrow (D+1)z=Dz+z=\bigg(\frac{1}{2}e^{t}-50\cos(2t)\bigg)+\frac{1}{2}e^{t}-25\sin(2t)\)
\(\Rightarrow (D+1)z=e^{t}-25\sin(2t)-50\cos(2t)\)
Putting \(Dx=e^{t}-y\) and the value of \((D+1)z\) in first equation, we get,
\(e^{t}-y+Dy+e^{t}-25\sin 2t-50\cos 2t=0\)
\(\Rightarrow Dy-y=25\sin 2t + 50\cos 2t-2e^{t}\)
So the integrating factor of this differential equation is,
\(I.F=e^{\int(-1)dt}\)
\(=e^{-t}\)
Hence the solution is,
\(ye^{-t}=\int(25\sin2t+50\cos2t-2e^{t})e^{-t}dt\)
\(=\int25\sin2te^{-t}dt+\int50\cos2te^{-t}dt-2\int e^{t}e^{-t}dt\)
\(=-10e^{-t}\cos2t-5e^{-t}\sin2t+20e^{-t}\sin2t-10e^{-t}\cos2t-2t+C\)
\(\Rightarrow y=-10\cos2t-5\sin2t+20\sin2t-10\cos2t-2te^{t}+Ce^{t}\)
\(\Rightarrow y=-20\cos2t+15\sin2t-2te^{t}+Ce^{t}\)
Now putting this value of \(y\) in the given second differential equation, we get: \(Dx=e^{t}-y\)
\(=e^{t}-(-20\cos2t+15\sin2t-2te^{t}+Ce^{t})\)
\(=e^{t}+20\cos2t-15\sin2t+2te^{t}-Ce^{t}\)
Integrating both sides, we get,
\(x=\int(e^{t}+20\cos2t-15\sin2t+2te^{t}-Ce^{t})dt\)
\(=e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}-2e^{t}-Ce^{t}+D\)
\(=(-C-1)e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D\)
\(=C'e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D\)
Hence the required solution is,
\(x=C'e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D\)
\(y=-20\cos2t+15\sin2t-2te^{t}+Ce^{t}\)
\(z=\frac{1}{2}e^{t}-25\sin2t\)
Note: For particular value \(C'=-\frac{1}{2} \text{ and } C=-\frac{1}{2}\) these solutions satisfies
the given system of differential equations.
The required solution is,
\(x=C'e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D\)
\(y=-20\cos2t+15\sin2t-2te^{t}+Ce^{t}\)
\(z=\frac{1}{2}e^{t}-25\sin2t\)
\(Dx+Dy+(D+1)z=0\)
\(Dx+y=e^{t}\)
\(Dx+y-2z=50\sin(2t)\)
Putting the value of \(Dx+y\) in the third equation, we get,
\(Dx+y-2z=50\sin(2t)\)
\(\Rightarrow e^{t}-2z=50\sin(2t)\)
\(\Rightarrow z=\frac{1}{2}e^{t}-25\sin(2t)\)
Now differentiating both sides with respect to t, we get,
\(Dz=\frac{1}{2}e^{t}-50\cos(2t)\)
\(\Rightarrow (D+1)z=Dz+z=\bigg(\frac{1}{2}e^{t}-50\cos(2t)\bigg)+\frac{1}{2}e^{t}-25\sin(2t)\)
\(\Rightarrow (D+1)z=e^{t}-25\sin(2t)-50\cos(2t)\)
Putting \(Dx=e^{t}-y\) and the value of \((D+1)z\) in first equation, we get,
\(e^{t}-y+Dy+e^{t}-25\sin 2t-50\cos 2t=0\)
\(\Rightarrow Dy-y=25\sin 2t + 50\cos 2t-2e^{t}\)
So the integrating factor of this differential equation is,
\(I.F=e^{\int(-1)dt}\)
\(=e^{-t}\)
Hence the solution is,
\(ye^{-t}=\int(25\sin2t+50\cos2t-2e^{t})e^{-t}dt\)
\(=\int25\sin2te^{-t}dt+\int50\cos2te^{-t}dt-2\int e^{t}e^{-t}dt\)
\(=-10e^{-t}\cos2t-5e^{-t}\sin2t+20e^{-t}\sin2t-10e^{-t}\cos2t-2t+C\)
\(\Rightarrow y=-10\cos2t-5\sin2t+20\sin2t-10\cos2t-2te^{t}+Ce^{t}\)
\(\Rightarrow y=-20\cos2t+15\sin2t-2te^{t}+Ce^{t}\)
Now putting this value of \(y\) in the given second differential equation, we get: \(Dx=e^{t}-y\)
\(=e^{t}-(-20\cos2t+15\sin2t-2te^{t}+Ce^{t})\)
\(=e^{t}+20\cos2t-15\sin2t+2te^{t}-Ce^{t}\)
Integrating both sides, we get,
\(x=\int(e^{t}+20\cos2t-15\sin2t+2te^{t}-Ce^{t})dt\)
\(=e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}-2e^{t}-Ce^{t}+D\)
\(=(-C-1)e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D\)
\(=C'e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D\)
Hence the required solution is,
\(x=C'e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D\)
\(y=-20\cos2t+15\sin2t-2te^{t}+Ce^{t}\)
\(z=\frac{1}{2}e^{t}-25\sin2t\)
Note: For particular value \(C'=-\frac{1}{2} \text{ and } C=-\frac{1}{2}\) these solutions satisfies
the given system of differential equations.
The required solution is,
\(x=C'e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D\)
\(y=-20\cos2t+15\sin2t-2te^{t}+Ce^{t}\)
\(z=\frac{1}{2}e^{t}-25\sin2t\)
