A. Let g(t) be the solution of the initial value problem 6t dy/dt+y=0, t>0, with

Question

A. Let g(t) be the solution of the initial value problem

6t dy/dt+y=0, t>0, with g(1)=1.
Find g(t).

g(t)=....................................... .

B. Let f(t) be the solution of the initial value problem

6t dy/dt+y= t^2 with f(0)=0.
Find f(t).

f(t)=........................... .

C. Find a constant c so that
k(t)=f(t)+cg(t)
solves the differential equation in part B and k(1)=3.
c= .

Explanation / Answer

A) 6t dy/dt +y = 0 6t dy/dt = -y => (1/y) dy = (-1/6t) dt integrate ln(y) = -ln(6t) +C ln(y) + ln(6t) = C ln(1) + ln(6) = C C = 1.7917 = ln(6) so ln(y) + ln(6t) =ln(6) ln(y 6t) = ln(6) y 6t = e^ln(6) = 6 yt = 1 y = 1/t g(t) = 1/t B) 6t dy/dt +y = t^2 dy/dt + (1/6t) y = t/6 Integrating Factor = e^integral (1/6t) dx = e^ln(t)^(1/6) = t^1/6 y . t^1/6 = integral ( t/6 . t^1/6) dt = integral (t^7/6 /6) dt = (1/6) t^(13/6) /(13/6) = (1/13) t^(13/6) +C y = (1/13) t^(2) + t^(-1/6) C y(0) = 0 0 = C hence f(t) =( 1/13) t^2 C) k(t)=f(t)+cg(t) => k(t) = (1/13) t^2 + C (1/t ) k(1) = 3 => 3 = (1/13) + C => C = 38/13

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