How can I solve the following initial value problem?

Question

We know that there's an amount of pollution that's being wasted in a day, and 30% of it disappears. We then assume that the amount of pollution released in a day is $M$ kg and that the lake is free from the waste in the start. Explain that amount substance $y(t)$ in the lake at time $t$, measured in a day is the following:

$$y'(t)= - \frac{3}{10} y(t) + M$$

We know that $y(0)=0$

How can I show that the substitution is possibly when $$u(t)=y(t)-\frac{10}{3}M$$

can be rewritten to $u'(t)=-3/10 u(t)$

and

$u(0)=-\frac{3}{10}M$

As well as solving the start value problem. I'm quite confused as how to start and where to go. Both $u(t)$ and $y(t)$ should be expressed with $M$

Answer 1

$U(t) = y(t) - \frac{10}{3}M$

Thus $y(t) = u(t) + \frac{10}{3}M$

Substitute this in the original pde

$y'(t) = u'(t)$

Thus $u'(t) = -\frac{3}{10}(u(t)) + \frac{10}{3}M) +M$

$u'(t) = -\frac{3}{10}(u(t)$ with $u(0) =y(0) - \frac{10}{3}M = -\frac{10}{3}M$

Put $u(t) = Ce^{-\frac{3}{10}t}$

$u(0) = -\frac{10}{3}M = C $

$u(t) = -\frac{10}{3}Me^{-\frac{3}{10}t}$

$y(t) =u(t) + \frac{10}{3}M$

$y(t) = -\frac{10}{3}Me^{-\frac{3}{10}t}+\frac{10}{3}M$$= \frac{10}{3}M\left(1-e^{-\frac{3}{10}t}\right)$

Answer 2

This equation is separable. How you factor the right side is up to taste, the given solution prefers that the variable factor has a coefficient of $1$ for $y$, $$ y' = -\frac{3}{10}\cdot\left(y-\frac{10}3M\right). $$ The separation of variables then proceeds to bring the factor with $y$ to the left side and then integrate, $$ \int\frac{dy}{y-\frac{10}3M}=-\frac{3}{10}\int\,dt. $$ To find the standard form for the integral on the left side you now substitute the denominator as new variable, $u=y-\frac{10}3M$ to find $$ -\frac{3}{10}t+c = \int\frac{du}{u}=\ln|u|=\ln\left|y-\frac{10}3M\right| $$ or $$ y=\frac{10}3M+C\,e^{-\frac{3}{10}t}. $$