separation of variables first order differential equation problem Hello I've hit a wall on this one. It's quite straightforward but my brain is jammed!
I want to find the general solution of:
$$\frac {dm}{dt} \,+\,\,\frac {5m}{50+2t}\,\,=\,\frac 75\,\,(t\ge0)$$
$$\,\,$$
This is a separation of variables problem since $$\frac {5m}{50+2t}$$
$$\,\,$$
is a division of the dependent variable by the independent variable (at least I think so)
$$\,\,$$
so rearranging yields:
$$\,\,$$
$$\frac {1}{5m}\frac {dm}{dt}\,\,-\,\,\frac {7}{25m}\,\,=\,\,-\frac {1}{50+2t}$$
$$\,\,$$
integrating both sides yields:
$$\int\left(\frac {1}{5m}\,\,-\,\,\frac {7}{25m}\right)\,\,dm\,\,=\,\,-\int\left(\frac {1}{50+2t}\right)\,\,dt$$
$$\,\,$$
$$-\int\frac {2}{25m} \,\, dm\,\,=\,\,-\int\left(\frac {1}{50+2t}\right)\,\,dt$$
$$\,\,$$
$$-\frac {2}{25}\int\frac {1}{m} \,\, dm\,\,=\,\,-\int\left(\frac {1}{50+2t}\right)\,\,dt$$
$$\,\,$$
$$-\frac {2}{25}\ln|m| = -\ln|50+2t|\,+\,c$$
I'm I on the right track or am I just blowing hot wind?
 A: As people have pointed out this is a linear first-order equation and not a separable one. To that effect let $$p(t)=e^{\int\frac{5}{50+2t}dt}$$ Then multiply both sides of the equation by $p(t)$. Note that this is equivalent to $$D_t\big [m(t)p(t) \big ]=\frac{7}{5}p(t)$$ We can now integrate both sides with respect to $t$, note that we use the FTC on the LHS. $$m(t)p(t)=\int\frac{7}{5}p(t)dt+C \dots(1)$$ Note that $$p(t)=e^{\frac{5}{2}\ln|50+2t|}=|50+2t|^{\frac{5}{2}}=(50+2t)^{\frac{5}{2}}$$Using this information you should be able to easily solve equation (1) for $m(t)$ which is your solution. $p(t)$ is called the integrating factor so look it up if you have not seen this method before.
A: $$\frac{\text{d}m(t)}{\text{d}t}+\frac{5m(t)}{50+2t}=\frac{7}{5}\Longleftrightarrow$$
$$m'(t)+\frac{5m(t)}{50+2t}=\frac{7}{5}\Longleftrightarrow$$
$$m'(t)+\frac{5m(t)}{2(t+25)}=\frac{7}{5}\Longleftrightarrow$$

Multiply both sides by $(t+25)^{\frac{5}{2}}$:

$$(t+25)^{\frac{5}{2}}\cdot m'(t)+\frac{5(t+25)^{\frac{3}{2}}}{2}\cdot m(t)=\frac{7}{5}\cdot(t+25)^{\frac{5}{2}}\Longleftrightarrow$$

Subsitute $\frac{5(t+25)^{\frac{3}{2}}}{2}=\frac{\text{d}}{\text{d}t}\left((t+25)^{\frac{5}{2}}\right)$:

$$(t+25)^{\frac{5}{2}}\cdot m'(t)+\frac{\text{d}}{\text{d}t}\left((t+25)^{\frac{5}{2}}\right)\cdot m(t)=\frac{7}{5}\cdot(t+25)^{\frac{5}{2}}\Longleftrightarrow$$

Apply the reverse product rule to the left-hand side:

$$\frac{\text{d}}{\text{d}t}\left((t+25)^{\frac{5}{2}}m(t)\right)=\frac{7}{5}\cdot(t+25)^{\frac{5}{2}}\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}t}\left((t+25)^{\frac{5}{2}}m(t)\right)\space\text{d}t=\int\frac{7}{5}\cdot(t+25)^{\frac{5}{2}}\space\text{d}t\Longleftrightarrow$$
$$(t+25)^{\frac{5}{2}}m(t)=\frac{2\left(t+25\right)^{\frac{7}{2}}}{5}+\text{C}\Longleftrightarrow$$
$$m(t)=\frac{2t}{5}+\frac{\text{C}}{(t+25)^{\frac{5}{2}}}+10$$
A: Notice, the given equation is in linear differential form: $\frac{dy}{dx}+P(x)y=Q(x)$,
$$\frac{dm}{dt}+\frac{5m}{50+2t}=\frac 75$$
multiplying both the sides by integration factor: $e^{\large \int \frac{5}{50+2t}\ dt}=e^{\ln(t+25)^{5/2}}=(t+25)^{5/2}$, 
$$(t+25)^{5/2}\frac{dm}{dt}+(t+25)^{5/2}\frac{5m}{50+2t}=\frac 75(t+25)^{5/2}$$ 
$$\frac{d}{dt}\left(m(t+25)^{5/2}\right)=\frac 75(t+25)^{5/2}$$
$$\int d\left(m(t+25)^{5/2}\right)=\frac 75(t+25)^{5/2}\ dt$$
$$m(t+25)^{5/2}=\frac 75\frac{(t+25)^{7/2}}{7/2}+C$$
$$m(t+25)^{5/2}=\frac 25(t+25)^{7/2}+C$$
or $$m=\frac {2t}{5}+C(t+25)^{-5/2}+10$$
