A scaled model of a bar bending uses plain numbers and the equation y'' - 2y' + 5y = 10eˣ, with…

Step 1. The particular integral. With $y_p = \lambda e^{x}$: $\lambda(1 - 2 + 5)e^{x} = 4\lambda e^{x} = 10e^{x}$. So $\lambda = 2.5$.

Step 2. The general solution. $y = e^{x}(A\cos 2x + B\sin 2x) + 2.5e^{x}$.

Step 3. $A$. $y(0) = A + 2.5 = 0$, so $A = -2.5$.

Step 4. $B$. Differentiate the complementary part and add $2.5e^{x}$. The cosine term has the factor $P = A + 2B$ and the sine term has the factor $Q = B - 2A$. So

At $x = 0$: $A + 2B + 2.5 = 1$. With $A = -2.5$ this gives $2B = 1$, so $B = 0.5$.

Step 5. Evaluate. $y\left(\dfrac{\pi}{2}\right) = e^{\pi/2}(2.5 \times 1 + 0 + 2.5) = 5e^{\pi/2} = 24.0524$.