How do you find the general solution of the ODE?

Question

Find the general solution of the ODE: $$\left\{\begin{array}{l}{\dot{x}=43x+5y} \\ {\dot{y}=3x+7y y}\end{array}\right.$$

How would you find the general solution of the ODE?

Answer 1

Hint: Use the substitution $z=x+y$ and note that $z’(t)=3z(t)$. Secondly the substitution $w=x-y$ yields: $w’(t)=w(t)$

Answer 2

I'd write two new variables $u$ and $v$ each of which is a generic linear combination of $x$ and $y$. (Actually, I would write $x = au+bv$ and $y = cu+dv$.)The equations, when you make this substitution, are uglier, but they contain an expression involving $a,b,c,d)$ in front of the $v$ term in the equation for $\dot{u}$ and another such expression in front of the $u$ term in the equation for $\dot{v}$.

So you can carefully choose $a,b,c$ and $d$ so that those two expressions both vanish and you are left with a pair of easy one-variable differential equations looking like $$ \dot{u} = g u \\ \dot{v} = h v $$ Those are easy to solve, and then you substitute to get $s$ and $y$.

Answer 3

The equation is $X' = MX$ where $X$ is the vector of functions $X=\binom{x}{y}$, and the matrix $M=\left( \begin{array}{ll}{2} & {1} \\ {1} & {2}\end{array}\right)$ can be written as $M=SJS^{-1}$, where $$S=\left( \begin{array}{cc}{-1} & {1} \\ {1} & {1}\end{array}\right),\quad J=\left( \begin{array}{ll}{1} & {0} \\ {0} & {3}\end{array}\right),\quad S^{-1}=\left( \begin{array}{cc}{-\frac{1}{2}} & {\frac{1}{2}} \\ {\frac{1}{2}} & {\frac{1}{2}}\end{array}\right).$$ With this, if you write $Y=S^{-1}X$, then $Y$ solves an ODE with a diagonal matrix (the "decoupled system") $$ Y' = J Y.$$ This is trivial to solve using 1D theory, and then you can recover $X$ as $X=SY$.

Answer 4

HINT....An alternative method not involving matrices is to differentiate one of the equations and then eliminate one variable. You then end up with a second order linear differential equation with constant coefficients, which is straightforward to solve.

So in this case, differentiating the first equation and eliminating $y$ and its derivative leads to $$\ddot{x}-4\dot{x}+3x=0$$

Solve this and then substitute back to get $y$