A. Let g(t) be the solution of the initial value problem 4t*dy/dt+y=0 t>0 with g

Question

A. Let g(t) be the solution of the initial value problem
4t*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
4t*dy/dt+y=t^3
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)=11
c=

Explanation / Answer

A. ) 4t dy/dt +y = 0 4t dy/dt = -y (1/y) dy = (-1/4t) dt integrate ln y = -(1/4) ln t + C y = e^((-1/4) ln t+C) g(t) = e^((-1/4) ln t+C) g(1) = 1 => 1 = e^(-1/4) ln(1) +C => ln(1) = (-1/4) ln(1)+C C = 5/4ln(1) = 0 hence g(t) = e^((-1/4) ln t B) 4t*dy/dt+y=t^3 dy/dt + (1/4t) y = t^2/4 IF = e^integral (1/4t) dt = e^1/4lnt = t^1/4 so y * t^(1/4) = integral ( t^(1/4) * t^2 /4 ) dt = integral (1/4) t^9/4 dt = (1/4)/(13/4) t^(13/4) +C y * t^(1/4) = (1/13) t^(13/4) +C f(0) = 0 => 0(0) = (1/13)(0) +C => C= 0 hence f(t) = (1/13) t^(13/4) / t(1/4) = (1/13) t^3 C) k(t)=f(t)+cg(t) k(t) = (1/13) t^3 + c e^((-1/4) ln t k(1) = 11 => 11 = (1/13) + C e^(0) C = 11-(1/13) = 142/13 C = 142/13 = 10.923

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