Solve the IVP system $\displaystyle{\begin{cases}x_1'=3x_1-4x_2+4x_3\\x_2'=4x_1-5x_2+4x_3\\x_3'=4x_1-4x_2+3x_3\\x_1(0)=2,\ x_2(0)=1,\ x_3(0)=-1\end{cases}}$
I am having trouble solving this. I know one method involves finding the eigenvalues and eigenvectors but is there not a method without using linear algebra and eigenvalues?
Hint
Solve this first then ...Substract 3) from 2)
$$x'_2-x'_3=-(x_2-x_3)$$ $$(x_2-x_3)'=-(x_2-x_3)$$
Substitute $z=x_2-x_3$
Then the equation becomes a simple first ODE easy to solve
$$z'=-z$$ $$\ln(z)=-t+K$$ $$z=Ke^{-t}$$ $$x_2(0)-x_3(0)=K \implies K=2 \implies x_2-x_3=2e^{-t}$$ First equation becomes ... $$ \begin{align} 1)x'_1&=3x_1-4x_2+4x_3\\ 1)x'_1&=3x_1-4z\\ 1)x'_1&=3x_1-8e^{-t}\\ \end{align} $$ Which is easy to solve ..
First equation ($x_1$)
$$x'_1=3x_1-8e^{-t}$$ $$x'_1-3x_1=-8e^{-t}$$ $e^{-3t}$ as integrating factor $$x'_1e^{-3t}-3x_1e^{-3t}=-8e^{-t}e^{-3t}$$ $$(x_1e^{-3t})'=-8e^{-4t}$$ Simply integrate now $$x_1e^{-3t}=-8\int e^{-4t}dx =2e^{-4t}+C$$ $$x_1=2e^{-t}+Ce^{3t}$$ We need to evaluate the constant C for $t=0$ $$x_1(0)=2+C \implies C=0 \implies x_1=2e^{-t}$$ $$\boxed{x_1=2e^{-t}}$$
Second Equation for $x_2$ $$x'_2=4x_1-5x_2+4x_3$$ We know the value of $x_1$ and we have a relation between $x_2$ and $x_3$
Because $x_1=2e^{-t}$ $$x'_2=8e^{-t}-5x_2+4x_3$$ Because $x_2-x_3=2e^{-t} \implies x_3= x_2-2e^{-t}$ $$x'_2=8e^{-t}-5x_2+4(x_2-2e^{-t})$$ $$x'_2=-x_2$$ $$x_2=Re^{-t} $$ Evaluate the constant R for $t=0$ $$x_2(0)=1 \implies R=1$$ $$\boxed {x_2=e^{-t} }$$
Third equation $x_3$
You don't need to solve any differential equation since you have a relation between $x_2$ and $x_3$ $$x_3=x_2-2e^{-t} =e^{-t}-2^{-t}$$ Therefore $$\boxed{x_3=-e^{-t}}$$ You have finished...