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Solve the boundary value problem
(d^(2)y)/(dx^(2)) + 4 (dy)/(dx) - 5y=0


y(0)=0

y( \pi /2)=1

User Noor Nawaz
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1 Answer

6 votes
Characteristic equation:


r^2+4r-5=(r+5)(r-1)=0\implies r=-5,r=1


\implies y_c=C_1e^(-5x)+C_2e^x

Given that
y(0)=0 and
y\left(\frac\pi2\right)=1, you have


\begin{cases}0=C_1+C_2\\1=C_1e^(-5\pi/2)+C_2e^(\pi/2)\end{cases}\implies C_1=(e^(5\pi/2))/(1-e^(3\pi)),C_2=-(e^(5\pi/2))/(1-e^(3\pi))

so that the particular solution is


y=(e^(5\pi/2))/(1-e^(3\pi))(e^(-5x)-e^x)
User Edd Morgan
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