First order ode... dr/dt=a/p-br/3p

Question

usukidoll · Answer

$\frac{dr}{dt}=\frac{a}{p}-\frac{b}{3p}r$\

Standard Linear ODE form:\

$\frac{dy}{dx}+p(x)y=q(x)$\
Since we have $y=r$ and $x=t$, we have\

$\frac{dr}{dt}+p(t)r=q(t)$\
Therefore, our $p(t) = \frac{b}{3p}$ and our $q(t) = \frac{a}{p}$\

Also, our $v(t) = e^{\int \frac{b}{3p}} \rightarrow e^{ \frac{b}{3p}t}$\

Multiplying our integrating factor throughout the equation, we have, \

$e^{\frac{b}{3p}t} (\frac{dr}{dt}) + e^{ \frac{b}{3p}t} (\frac{b}{3p}) =e^{\frac{b}{3p}t}(\frac{a}{p})$\

Using reverse product rule, we have,\

$r (e^{ \frac{b}{3p}t}) =\int e^{ \frac{b}{3p}t}(\frac{a}{p})$\

Integrating the right side, we have\
$r (e^{ \frac{b}{3p}t}) =\frac{3a}{b}(e^{ \frac{b}{3p}t})+C $\

Dividing $(e^{ \frac{b}{3p}t})$ throughout the whole equation, we have,\

$r  =\frac{3a}{b}+\frac{C}{(e^{ \frac{b}{3p}t})} $\

$r  =\frac{3a}{b}+C((e^{- \frac{b}{3p}t})) $\

Now, suppose our Initial Condition is $r(0)=0$. Then we have,\

$r(0)  =\frac{3a}{b}+C((e^{- \frac{b}{3p}(0)})) $\

$0  =\frac{3a}{b}+C(e^{0}) $\

$0  =\frac{3a}{b}+C $\

$-\frac{3a}{b}=C $\

Substituting C back into the equation, we have, \
$r  =\frac{3a}{b}-\frac{3a}{b}((e^{- \frac{b}{3p}t})) $\

usukidoll · Answer

I think the fact that using separation of variables is impossible on this due to the fact that it's not in the form of h(y) dy =f(x) dx . Our equation is $\frac{dr}{dt}=\frac{a}{p}-\frac{b}{3p}r$\. We can only use separation of variables with multiplication or division, but never with addition or subtraction.

usukidoll · Answer

like if our equation was $$/\frac{dx}{dy} = x$$ we can use separation of variables, but the equation I was given breaks the rules real fast.

usukidoll · Answer

you can't have something like $$\frac{dy}{dx} = x+y $$

usukidoll · Answer

and expect separation of variables to work.. you can't add  to both sides or subtract both sides XD

ganeshie8 · Answer

dr/dt=a/p-br/3p

is separable right ?

usukidoll · Answer

I don't think so? Not unless the r is everywhere on the right side

usukidoll · Answer

I have this equation |dw:1421914368960:dw| there is no () around