Solve this linear differential equation.?

Question

$$S'(t) = 2 - \frac {S(t) }{(100 + t)}$$

Assuming $100 + t$ is positive

I got the integrating factor is $e^{\ln(100 +t)}$

Then $$s(t) (100 + t) = \int 2(100 +t)\ dt$$

So $$s(t) = {200t +t^2 \over 100 + t} + {c \over 100+t}$$

But my book says it's $$s(t) = {100 + t} + {c \over 100+t}$$

Answer 1

If you mean this $$S'(t) = 2 - \frac {s(t) }{ (100 + t)}$$ $$100S'(t)+S't +S= 200+2t $$ $$100S'(t)+(tS(t))'= 200+2t $$ Integrate $$100S(t)+tS(t)= 200t+t^2+K $$ $$S(t)= 100+t+\frac K {100+t}$$

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Edit

Your answer is correct...

Both answers are the same

$$s(t) = {200t +t^2 \over 100 + t} + {c \over 100+t}$$ $$s(t) = {100t+ 100t +t^2 \over 100 + t} + {c \over 100+t}$$ $$s(t) =t+ {100t \over 100 + t} + {c \over 100+t}$$ $$s(t) =t+ {100t +c\over 100 + t} $$ Substitute $c=10 000+K$ it's just a constant $$s(t) =t+ {100t +10000+K\over 100 + t} $$ $$s(t) =t+ 100+{K\over 100 + t} $$

Answer 2

Yep: your answer is correct:

$$S(t) = \frac{c_1}{t+100}+\frac{t^2+200 t}{t+100}$$