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given that y = 2 at x = 0 and [itex] \frac{dy}{dx} = -5 [/itex] at x = 0, find y in terms of x given further that
[tex] \frac{d^2y}{dx^2} + \frac{dy}{dx} = 2x +3 [/tex]
finding the complementary function:
m^2 + m = 0
m(m+1) = 0
m = 0, m = -1
so complementary function y = A + Be^(-x)
Particular integral:
let y = px + q
[itex] \frac{dy}{dx} = p [/itex]
[itex] \frac{d^2y}{dx^2} = 0 [/itex]
so subbing this into the second order DE: 0 + p = 2x + 3
hence p = 3
so particular integral is y = 3x + q
hence general solution is:
y = A + Be^(-x) + 3x + q
y = 2, x = 0
hence 2 = A + B + q
dy/dx = -Be^(-x) + 3
dy/dx = -2 at x = 0 hence B = 8
so 2 = A + 8 + q
how do I find A and q?
[tex] \frac{d^2y}{dx^2} + \frac{dy}{dx} = 2x +3 [/tex]
finding the complementary function:
m^2 + m = 0
m(m+1) = 0
m = 0, m = -1
so complementary function y = A + Be^(-x)
Particular integral:
let y = px + q
[itex] \frac{dy}{dx} = p [/itex]
[itex] \frac{d^2y}{dx^2} = 0 [/itex]
so subbing this into the second order DE: 0 + p = 2x + 3
hence p = 3
so particular integral is y = 3x + q
hence general solution is:
y = A + Be^(-x) + 3x + q
y = 2, x = 0
hence 2 = A + B + q
dy/dx = -Be^(-x) + 3
dy/dx = -2 at x = 0 hence B = 8
so 2 = A + 8 + q
how do I find A and q?