Find the particular solution of the differential equation dy/dx+6y=5 satisfying

Question

Find the particular solution of the differential equation dy/dx+6y=5 satisfying the initial condition y(0)=0. Answer: y= . Your answer should be a function of x.

Explanation / Answer

a) This can be solved by separation of variables method [1] dy/dx = f(x)·g(y) 1/g(y) dy = f(x) => ? 1/g(y) dy = ? f(x) dx + c dy/dx + 4·y = 5 dy/dx = 5 - 4·y = - 4·(y - (5/4)) 1/(y - (5/4)) dy = - 4 dx ? 1/(y - (5/4)) dy = ? - 4 dx + c ln(y - (5/4)) = c - 4·x y - (5/4) = e^(c - 4·x) = e^(c) · e^(-4·x) With C = e^(c) the general solution is given by: y(x) = (5/4) + C·e^(-4·x) Apply initial condition to evaluate constant C: y(0) = 0 (5/4) + C·e^(-4·0) = 0 (5/4) + C = 0 C = (-5/4) So the solution to this initial value problem is: y(x) = (5/4)·(1 - e^(-4·x)) b) This is non-separable 1st order ODE. It can be made separable and solved by integrating factor method [2]. The integrating factor for an ODE of the form: (dy/dx) + p(x)·y(x) = q(x) is µ(x) = e^( ? p(x) dx ) and leads to µ(x)·(dy/dx) + µ(x)·p(x)·y(x) = µ(x)·q(x) µ(x)·(dy/dx) + (dµ/dx)·y(x) = µ(x)·q(x) d(µ·y)/dx = µ(x)·q(x) => µ(x)·y(x) = ? µ(x)·q(x) dx + C => y(x) = [? µ(x)·q(x) dx + C]/µ(x) f'(t) - f(t) = - 2·t => p(t) = -1 q(t) = -2·t => µ(t) = e^( ? -1 dt ) = e^(-t) => f(t) = [? e^(-t)·(-2·t) dt + C]/e^(-t) = [? e^(-t)·(-2·t) dt + C] · e^(t) = [ 2·e^(-t)·(t + 1) + C] · e^(t) = 2·(t + 1) + C·e^(t) Apply initial condition y(1) = 1 2·(1 + 1) + C·e^(1) = 1 C = -3/e = -3·e^(-1) => f(t) = 2·(t + 1) - 3·e^(-1)·e^(t) = 2·(t + 1) - 3·e^(t - 1)

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