Question

In Problems 47-50 use systematic elimination to solve the given system.

$\begin{array}{c}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{t}\mathbf{-}\mathbf{2}\\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{t}\end{array}$

Short answer

The solution is$x\left(t\right)=c_1{e}^t+c_2{e}^{5t}-\frac{8}{5}t-\frac{3}{25}\text{and}y\left(t\right)=-c_1{e}^t+3c_2{e}^{5t}+\frac{21}{5}t-\frac{84}{25}$

Step-by-step solution

Given Data

Differential equation is the equation which relate one or more unknown functions and their derivatives.

Given differential equation,

$\begin{array}{c}\frac{dx}{dt}=2x+y+t-2\\ \frac{dy}{dt}=3x+4y-4t\end{array}$

Find differential operating

We have the system of first order equations

$\begin{array}{rr}& \frac{\mathrm{dx}}{\mathrm{dt}}=2\mathrm{x}+\mathrm{y}+\mathrm{t}-2\\ & \frac{\mathrm{dy}}{\mathrm{dt}}=3\mathrm{x}+4\mathrm{y}-4\mathrm{t}\end{array}$

which by using differential operating is

$\begin{array}{rr}& Dx-2x-y=t-2\\ & Dy-3x-4y=-4t\end{array}$

which is

$\begin{array}{c}(D-2)x-y=t-2\\ -3x+(D-4)y=-4t\end{array}$

and we have to solve it using systematic elimination as the following technique:
Operate the first equation by the operator$\left(D-4\right)$ and multiply the second equation by the number 1, then we obtain

$\begin{array}{c}(D-2)(D-4)x-(D-4)y=(D-4)(t-2)\\ -3x+(D-4)y=-4t\end{array}$

After that, add equation $\left(1\alpha \right)$to equation $\left(2\alpha \right)$and simplify, then we obtain

$\begin{array}{c}[(D-2)(D-4)x-(D-4)y]+[-3x+(D-4)y]=(D-4)(t-2)-4t\\ [(D-4)y-(D-4)y]+[(D-2)(D-4)x+3x]=D(t-2)-4(t-2)-4t\\ (D-2)(D-4)x-3x=1-4t+8-4t\\ (D^2-6D+8)x-3x=-8t+9\\ (D^2-6D+5)x=-8t+9\end{array}$

Simplify homogeneous equation

After that, we have to obtain the solution of the corresponding homogeneous equation$(D^2-6D+5)x=0$ as the following:
We have the auxiliary solution for $x$as

$\begin{array}{c}(m^2-6m+5)=0\\ (m-1)(m-5)=0\end{array}$

Then we have the roots

$m_1=1,m_2=5$

which are real and distinct.
Then we can obtain the homogeneous solution for $x$as

$x_c=c_1{e}^t+c_2{e}^{5t}$

After that, we have to obtain the particular solution by assuming that $x_p=At+B$, then differentiate with respect to $t$as

$\begin{array}{rr}& D{x}_p=A\\ & D^2{x}_p=0\end{array}$

Then substitute with$x_p=At+B$and equations$\left(5\right)$and (6) into equation (3), then we obtain

$\begin{array}{c}0-6A+5(At+B)=-8t+9\\ 5At+(5B-6A)=-8t+9\end{array}$

Then by comparison we can have

$5A=-8\text{then}:A=-\frac{8}{5}$

and$5B-6A=9$ then$:B=-3/25$
Then we can obtain the particular solution for$x$ as

$x_p=-\frac{8}{5}t-\frac{3}{25}$

Then from equation $\left(4\right)$and $\left(7\right)$, we can obtain the general solution for $x$as

$\begin{array}{rl}x\left(t\right)& =x_c+x_p\\ & =c_1{e}^t+c_2{e}^{5t}-\frac{8}{5}t-\frac{3}{25}\end{array}$

Final calculation and general solution

Now we have to obtain the general solution for$y$ by substituting with the solution of $x$shown in equation (8) into equation (1) as

role="math" localid="1667903494388" $\begin{array}{rcl}(D-2)\left(c_1{e}^t+c_2{e}^{5t}-\frac{8}{5}t-\frac{3}{25}\right)-y& =& t-2\\ D\left(c_1{e}^t+c_2{e}^{5t}-\frac{8}{5}t-\frac{3}{25}\right)-2\left(c_1{e}^t+c_2{e}^{5t}-\frac{8}{5}t-\frac{3}{25}\right)-y& =& t-2\\ \left(c_1{e}^t+5c_2{e}^{5t}-\frac{8}{5}\right)-\left(2c_1{e}^t+2c_2{e}^{5t}-\frac{16}{5}t-\frac{6}{25}\right)-y& =& t-2\end{array}$

Then we have

$\begin{array}{cc}y\left(t\right)& =(t-2)+(c_1{e}^t+5c_2{e}^{5t}-\frac{8}{5})-\left(2c_1{e}^t+2c_2{e}^{5t}-\frac{16}{5}t-\frac{6}{25}\right)\\ & =-c_1{e}^t+3c_2{e}^{5t}+\frac{21}{5}t-\frac{84}{25}\end{array}$

is the general solution for$y$ .

Hence, the solution is $x\left(t\right)=c_1{e}^t+c_2{e}^{5t}-\frac{8}{5}t-\frac{3}{25}\text{and}y\left(t\right)=-c_1{e}^t+3c_2{e}^{5t}+\frac{21}{5}t-\frac{84}{25}$.