Solve the boundary value problem y'' + y = 1 , with y(0)=0…

Solve the boundary value problem $y'' + y = 1$, with $y(0)=0, \qquad y(\pi/2)=0$.

Step 1: Homogeneous solution

\[ r^2+1=0 \implies r=\pm i \implies Y_h = C_1\cos x + C_2\sin x. \]

Step 2: Particular solution

The right-hand side is a constant, so try $y^*=A$: \[0 + A = 1 \implies A=1, \qquad y^*=1.\]

Step 3: General solution

\[ y = C_1\cos x + C_2\sin x + 1. \]

Step 4: Apply the boundary conditions

From $y(0)=0$: $C_1 + 1 = 0 \implies C_1 = -1.$

From $y(\pi/2)=0$: $C_1\cdot 0 + C_2\cdot 1 + 1 = 0 \implies C_2 = -1.$

Step 5: Verify

With $y=1-\cos x-\sin x$: $y'=\sin x-\cos x$, $y''=\cos x+\sin x$, so $y''+y=\cos x+\sin x+1-\cos x-\sin x=1. $\checkmark$$ Also $y(0)=1-1-0=0\ \checkmark$ and $y(\pi/2)=1-0-1=0. $\checkmark$$

Step 6: Solution

The solution is recorded in the result below.