Answer:
[tex]y(x) = 0[/tex]
Step-by-step explanation:
Let P(1) = 0, the differential equation becomes:
y" + 3y' + 2y = 0
Also let y' = my and y'' = m²y
Substitute
m²y+3my+2y = 0
(m²+3m+2)y = 0
The auxiliary equation is expressed as:
m²+3m+2 = 0
Factorize
m²+2m+m+2 = 0
m(m+2)+1(m+2) = 0
(m+1)(m+2) = 0
m+1 = 0 and m+2 = 0
m₁ = -1 and m₂ = -2
Since the value of m is real and distinct, the solution to the differential equation will be expressed as:
[tex]y = C_1e^{m_1x}+C_2e^{m_2x}\\[/tex]
[tex]y(x) = C_1e^{-x}+C_2e^{-2x}\\[/tex]
Given the initial condition y(0) = 0
[tex]y(0) = C_1e^{-0}+C_2e^{-2(0)}\\0 = C_1+C_2\\C_1+C_2 = 0 ................. 1[/tex]
If y'(0) = 0
[tex]y'(x) = -C_1e^{-x}-2C_2e^{-2x}\\y'(0) = -C_1e^{-0}-2C_2e^{-2(0)}\\0 = -C_1-2C_2\\C_1-2C_2 = 0...................... 2[/tex]
Solve eqn 1 and 2 simultaneously
From 1:
C₁ = -C₂ .......... 3
Substitute equation 3 into 2:
-C₂-2C₂ = 0
-3C₂ = 0
C₂ = 0
Substitute the constant into the differential equation.
[tex]y(x) = C_1e^{-x}+C_2e^{-2x}\\\\y(x) = 0[/tex]
