Solve the system x″ + y″ = t², x″ − y″ = 4t with x(0) = 8, x′(0) = 0, y(0) = 0, y′(0) = 0.

Step-by-step solution

1. 1

   Decouple the system by adding and subtracting. The two equations involve only the sum and the difference of $x''$ and $y''$, so linear combinations separate them immediately — no Laplace transform or matrix method is needed:

   $\text{add: } 2x'' = t^2 + 4t, \qquad \text{subtract: } 2y'' = t^2 - 4t$

   Hence

   $x'' = \tfrac{1}{2}t^2 + 2t, \qquad y'' = \tfrac{1}{2}t^2 - 2t$

2. 2

   Integrate x″ once and apply x′(0) = 0.

   $x' = \frac{t^3}{6} + t^2 + C_1$

   Setting $t = 0$ gives $0 = C_1$, so

   $x' = \frac{t^3}{6} + t^2$

   Applying each condition as soon as its constant appears keeps the two constants from getting mixed up.

3. 3

   Integrate again and apply x(0) = 8.

   $x = \frac{t^4}{24} + \frac{t^3}{3} + C_2, \qquad C_2 = 8$

   $x(t) = \frac{t^4}{24} + \frac{t^3}{3} + 8$

4. 4

   Repeat for y with both initial values zero.

   $y' = \frac{t^3}{6} - t^2 + C_3, \qquad y'(0) = 0 \Rightarrow C_3 = 0$

   $y = \frac{t^4}{24} - \frac{t^3}{3} + C_4, \qquad y(0) = 0 \Rightarrow C_4 = 0$

   $y(t) = \frac{t^4}{24} - \frac{t^3}{3}$

   Only the sign of the cubic term and the constant $8$ distinguish $y$ from $x$ — exactly the asymmetry introduced by the $\pm 4t$.

5. 5

   Verify both differential equations and all four initial conditions. Differentiating twice:

   $x'' = \frac{t^2}{2} + 2t, \qquad y'' = \frac{t^2}{2} - 2t$

   $x'' + y'' = t^2 \ \checkmark, \qquad x'' - y'' = 4t \ \checkmark$

   At $t = 0$: $x = 8$, $x' = 0$, $y = 0$, $y' = 0$ — all four match. Note that $x - y = \frac{2t^3}{3} + 8$ and $x + y = \frac{t^4}{12} + 8$, the two decoupled combinations.