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Content: Reducing them to various first order differential equations in the normal form.

begin{cases} x' +x - y' = -t \ x' + y' + y = 1 end{cases}

My solution is this right? could anyone improve this for me?

$$ x = -t + y' - x'$$
$$ y = 1 - y' - x'$$
$$left(begin{array}{c}x\yend{array}right) = left(begin{array}{cc}-1 &1\-1&-1end{array}right)left(begin{array}{c}x'\y'end{array}right) + left(begin{array}{c}-t\1end{array}right)$$,

$$ V = A cdot V'$$
$$ V' = A^{-1} cdot V - b $$

$$left(begin{array}{c}x'\y'end{array}right) = left(begin{array}{cc}-frac{1}{2} &frac{1}{2}\-frac{1}{2}&-frac{1}{2}end{array}right)left(begin{array}{c}x\yend{array}right) + left(begin{array}{c}t\-1end{array}right)$$

$$V_{p}(t) = left(begin{array}{c}x\yend{array}right)t + left(begin{array}{c}a\bend{array}right)$$

$$left(begin{array}{c}t\-1end{array}right) = V'_{p}(t) -left(begin{array}{cc}-frac{1}{2} &frac{1}{2}\-frac{1}{2}&-frac{1}{2}end{array}right) V_{p}(t)$$

$$frac{partial Biggr[ left(begin{array}{c}x\yend{array}right)t + left(begin{array}{c}a\bend{array}right) Biggr] }{partial t} - left(begin{array}{cc}-frac{1}{2} &frac{1}{2}\-frac{1}{2}&-frac{1}{2}end{array}right) Biggr[ left(begin{array}{c}x\yend{array}right)t + left(begin{array}{c}a\bend{array}right) Biggr]$$

$$left(begin{array}{c}x\yend{array}right) - Biggr[left(begin{array}{cc}-frac{x}{2} - frac{y}{2}\frac{x}{2}-frac{y}{2}end{array}right)t + left(begin{array}{cc}frac{-a}{2} - frac{b}{2}\frac{a}{2}-frac{b}{2}end{array}right) Biggr]$$

$$left(begin{array}{c}1\0end{array}right)t + left(begin{array}{c}0\-1end{array}right) = left(begin{array}{cc}frac{x}{2} + frac{y}{2}\-frac{x}{2}+frac{y}{2}end{array}right) + left(begin{array}{cc}x+frac{a}{2}+ frac{b}{2}\y-frac{a}{2}+frac{b}{2}end{array}right) $$

$$begin{cases} frac{x}{2} + frac{y}{2} = 1 \ -frac{x}{2} + frac{y}{2} = 0 end{cases}$$

$$begin{cases} x = 1 \ y = 1 end{cases}$$

$$begin{cases} x+frac{a}{2}+ frac{b}{2} = 0 \ y-frac{a}{2}+frac{b}{2} = -1 end{cases}$$

$$begin{cases} a = 1 \ b = -3 end{cases}$$

$$V_{p}t = left(begin{array}{c}1\1end{array}right)t + left(begin{array}{c}1\-3end{array}right)$$

$$V' = left(begin{array}{cc}-frac{1}{2} &-frac{1}{2}\frac{1}{2}&-frac{1}{2}end{array}right) $$

$$ beta = left(begin{array}{cc}-frac{1}{2} &-frac{1}{2}\frac{1}{2}&-frac{1}{2}end{array}right) $$

$$ V' = beta V $$

$$frac{dV}{V} = beta dt $$

$$ ln{|V|} = beta t + C $$
$$ |V| = e^{beta t} e^{C}$$
$$ V pm e^{C} e^{beta t}$$
$$ V_{h} = ke^{beta t} $$

$$ V = V_{h} + V_{p}$$
$$ V = ke^{beta t} + Biggr[ left(begin{array}{c}1\1end{array}right)t + left(begin{array}{c}1\-3end{array}right)Biggr]$$

$$k in Re$$

The normal form of a system of ODEs ist $boldsymbol{x}' = boldsymbol{f}(t,boldsymbol{x})$. Defining $boldsymbol{x} := (x,y)^{top}$ the given equations can be written in the form
begin{equation}
boldsymbol{underline{A}} boldsymbol{x}' + boldsymbol{x} = boldsymbol{b}, quad textrm{where} quad boldsymbol{underline{A}} := left( begin{array}{cc}
1 & -1\
1 & 1
end{array}
right) quad textrm{and} quad boldsymbol{b}(t) := left( begin{array}{c}
-t\
1
end{array}
right).
end{equation}
Because the matrix $boldsymbol{underline{A}}$ is invertible ($det(boldsymbol{underline{A}}) = 2 neq 0$) we may solve for the derivative:
begin{equation}
boldsymbol{x}' = boldsymbol{underline{A}}^{-1}left( boldsymbol{b} - boldsymbol{x} right) =: boldsymbol{f}(t,boldsymbol{x}), quad textrm{where } quad boldsymbol{underline{A}}^{-1} = frac{1}{2} left( begin{array}{cc}
1 & 1\
-1 & 1
end{array}
right).
end{equation}
In your answer you forgot to multiply $boldsymbol{b}$ by $boldsymbol{underline{A}}^{-1}$.

Now we have a non-homogeneous linear system of ODEs with constant coefficients, which may be solved using a matrix exponential. But I think the exercise was not to solve the equations, but just to reduce them to the normal form.

For a hint on the solution see Moo's comment.

Adding two equations gives
$$ x'=-frac12 x-frac12 y-frac12t+frac12. $$
Using the second equation to subtract the first one gives
$$ y'=frac12x-frac12y+frac12t+frac12. $$
Thus
$$ binom{x'}{y'}=left(begin{matrix}-frac12&-frac12\
frac12&-frac12
end{matrix}right)binom{x}{y}+binom{-frac12t+frac12}{frac12t+frac12}.$$