Solve the system of differential equations: (d^2y)/(dx^2) =…

Solve the system of differential equations: $$ \frac{d^2y}{dx^2} = z,\qquad \frac{d^2z}{dx^2} = y $$

Explanation of Method

Adding and subtracting the two equations decouples them into two independent second-order ODEs in $u=y+z$ and $v=y-z$, each solved by its own characteristic equation.

Define $u=y+z$, $v=y-z$ and add/subtract the equations.
Solve $u''=u$ and $v''=-v$ separately.
Recover $y=(u+v)/2$ and $z=(u-v)/2$.
Verify by substitution.

Step 1 · Decouple

With $u=y+z$, $v=y-z$, adding and subtracting the two equations gives

$$ u''-u=0,\qquad v''+v=0. $$

Step 2 · Solve each equation

$$ u(x)=A_1e^{x}+A_2e^{-x},\qquad v(x)=B_1\cos x+B_2\sin x. $$

Step 3 · Recover $y$ and $z$

Since $y=(u+v)/2$, $z=(u-v)/2$, absorbing the factor $\tfrac12$ into new constants gives

$$ y(x)=C_1e^{x}+C_2e^{-x}+C_3\cos x+C_4\sin x,\quad z(x)=C_1e^{x}+C_2e^{-x}-C_3\cos x-C_4\sin x. $$

Step 4 · Verification

Differentiating $y$ and $z$ twice reproduces $y''=z$ and $z''=y$ exactly. ✓

Summary Table

Quantity	Value
Sum equation	$u''-u=0\ \Rightarrow\ u=A_1e^x+A_2e^{-x}$
Difference equation	$v''+v=0\ \Rightarrow\ v=B_1\cos x+B_2\sin x$
Recovery	$y=(u+v)/2,\ z=(u-v)/2$

Additional Notes

The sum $u=y+z$ behaves like a hyperbolic oscillator, while the difference $v=y-z$ behaves like a harmonic oscillator with opposite sign.
This addition/subtraction trick works whenever a coupled system is symmetric under swapping the two variables.
Verified by substituting the closed forms back into both original second-order equations.

Result

The general solution of the system is:

$$ \begin{aligned} y(x) &= C_1 e^{x} + C_2 e^{-x} + C_3\cos x + C_4\sin x,\\ z(x) &= C_1 e^{x} + C_2 e^{-x} - C_3\cos x - C_4\sin x. \end{aligned} $$