Question

Solve the following boundary value problems.

(i) 𝑦

′′ + 4𝑦 = 0; 𝑦(0) = 3, 𝑦(𝜋/2) = −3,

(ii) 𝑦

′′ − 25𝑦 = 0; 𝑦(−2) = 𝑦(2) = cosh 10.

(iii) 𝑦

′′ + 2𝑦

′ + 2𝑦 = 0; 𝑦(0) = 1, 𝑦(𝜋/2) = 0.

Answer 1

• The associated characteristic equation is

$\lambda^2+4=0$ which gives the general solution $y= A\cos(2x)+B\sin(2x)$. Now, applying the boundary conditions gives us : $A=3$ and $B$ arbitrary (as $\sin(0) =\sin(\pi)=0$ ).

$y=3\cos(2x)+B\sin(2x)$

• The associated characteristic equation is

$\lambda^2-25=0$ which gives the general solution $Ae^{5x}+Be^{-5x}=A'\cosh(5x)+B'\sinh(5x)$. Applying the boundary conditions yields $A'=1, B'=0$, so $y=\cosh(5x)$.

• The associated characteristic equation is

$\lambda^2+2\lambda+2=0$ which gives the general solution $e^{-x}(A\sin(x)+B\cos(x))$. Applying the boundary conditions gives $B=1, A=0$, so the solution is $y=e^{-x}\cos(x)$