What is the solution for q^'+q/(20+t)=1,q(0)=5 ?

The solution for q^'+q/(20+t)=1,q(0)=5 is q(t)=(20t)/(20+t)+(t^2)/(2(20+t))+(100)/(20+t)

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q^'+q/(20+t)=1,q(0)=5

q^{^{'}} + \frac{q}{20 + t} = 1, q (0) = 5

q (t) = \frac{20 t}{20 + t} + \frac{t ^{2}}{2 ( 20 + t )} + \frac{100}{20 + t}

Solution steps

q^{'} (t) + \frac{q ( t )}{20 + t} = 1

Solve linear ODE:

q (t) = \frac{20 t}{20 + t} + \frac{t ^{2}}{2 ( 20 + t )} + \frac{100}{20 + t}

q (t) = \frac{20 t}{20 + t} + \frac{t ^{2}}{2 ( 20 + t )} + \frac{100}{20 + t}

x^{8} (x - 3)^{7}

n = 0 \sum \infty \frac{5}{4 ^{n + 1}}

x \to o lim (\frac{sin ( 4 x )}{x})

f^{^{'}} (x) = \frac{1}{x}

f (x) = 1 + 3 x

What is the solution for q^'+q/(20+t)=1,q(0)=5 ?
The solution for q^'+q/(20+t)=1,q(0)=5 is q(t)=(20t)/(20+t)+(t^2)/(2(20+t))+(100)/(20+t)