# Solve the given system of differential equations. [Dx+Dy+(D+1)z=0 Dx+y=e^{t} Dx+y-2z=50sin(2t)

Question
Differential equations
Solve the given system of differential equations.
\[Dx+Dy+(D+1)z=0\)
Dx+y=e^{t}\)
Dx+y-2z=50\sin(2t)\)

2021-02-22
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,
$$x=C'e^{t}+10\sin2t+\frac{15}{2}\cos2t+2te^{t}+D$$
$$y=-20\cos2t+15\sin2t-2te^{t}+Ce^{t}$$
$$z=\frac{1}{2}e^{t}-25\sin2t$$

### Relevant Questions

Solve the Differential equations
$$\displaystyle{\frac{{{\left({d}^{{{2}}}\right)}{y}}}{\rbrace}}{\left\lbrace{d}{\left({t}^{{{2}}}\right\rbrace}+{4}{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}+{3}{y}={e}^{{-{t}}}\right.}\right.}$$
Find the differential dy for the given values of x and dx. $$y=\frac{e^x}{10},x=0,dx=0.1$$
Solve the differential equations
(1) $$\displaystyle{x}{y}'-{2}{y}={x}^{{3}}{e}^{{x}}$$
(2) $$\displaystyle{\left({2}{y}{\left.{d}{x}\right.}+{\left.{d}{y}\right.}\right)}{e}^{{2}}{x}={0}$$
The coefficient matrix for a system of linear differential equations of the form $$\displaystyle{y}^{{{1}}}={A}_{{{y}}}$$ has the given eigenvalues and eigenspace bases. Find the general solution for the system.
$$\displaystyle{\left[\lambda_{{{1}}}=-{1}\Rightarrow\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}{0}{3}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},\lambda_{{{2}}}={3}{i}\Rightarrow\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}-{i}{1}+{i}{7}{i}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},\lambda_{{3}}=-{3}{i}\Rightarrow\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}+{i}{1}-{i}-{7}{i}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}\right]}$$
Solve
$$\left(d^{2}\frac{y}{dt^{2}}\right)\ +\ 7\left(\frac{dy}{dt}\right)\ +\ 10y=4te^{-3}t$$ with
$$y(0)=0,\ y'(0)=\ -1$$
Solve the Differential equations $$\displaystyle{\left({D}^{{3}}−{3}{D}+{2}\right)}{y}={0}$$
A particle moves along the curve $$\displaystyle{x}={2}{t}^{{2}}{y}={t}^{{2}}-{4}{t}$$ and z=3t-5 where t is the time.find the components of the velocity at t=1 in the direction i-3j+2k
Solve. $$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+\frac{{3}}{{x}}{y}={27}{y}^{{\frac{{1}}{{3}}}}{1}{n}{\left({x}\right)},{x}{>}{0}$$
Make and solve the given equation $$x\ dx\ +\ y\ dy=a^{2}\frac{x\ dy\ -\ y\ dx}{x^{2}\ +\ y^{2}}$$
Make and solve the given equation $$d^{3}\ \frac{y}{dx^{3}}\ -\ d^{2}\ \frac{y}{dx^{2}}=2x\ +\ 3$$