Compound interest Differential Equation A college student starts a savings account with an initial balance of $\$0$. He plans to save money at a continuous rate of $\$200$ per week. Also, at every week he plans to increase this rate by $\$10$. (ex. At the 4th month he would be saving at a rate of $\$240$ per week). Additionally, the college student finds a bank account that pays continuously compounded interest at a rate of $4\%$ per year. Estimate the time it'll take for the college student to save $\$500,000$. 
Hint: set up and solve a differential equation and plot the solution to make the final estimate.

My attempt:

The differential equation is hard to set up.

Let $S =$ amount saved.

Let $t =$ time.

$$\frac{dS}{dt} = \frac{0.04}{52}(200 + 10t)$$

I tried this differential equation but it doesn't satisfy the initial condition. 

Can someone help me come up with the differential equation?

Thanks!
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$$
\mbox{Let}\quad
\left\{\begin{array}{rclcl}
b_{n} &amp;:&amp; \mbox{Balance after}\ n\ \mbox{weeks}.&amp;&amp; b_{0} = 0
\\[2mm]
s_{0} &amp; : &amp; \mbox{Initial week saving} &amp; = &amp; 200
\\[2mm]
\Delta s &amp; : &amp; \mbox{Amount added to the every week saving} &amp; = &amp; 10
\\[2mm]
r &amp; : &amp; \mbox{Bank interest} \pars{~\mbox{per one per week}~}
&amp; = &amp; {4/\pars{12\times 4} \over 100} = {1 \over 1200}
\end{array}\right.
$$
We assumed $4$ weeks per month.


  $$
\begin{array}{rclc}
b_{0} &amp; = &amp; 0
\\
b_{1} &amp; = &amp; s_{0}
\\
b_{2} &amp; = &amp; b_{1}\pars{1 + r} + \pars{s_{0}  + \Delta s}
\\
b_{3} &amp; = &amp; b_{2}\pars{1 + r} + \pars{s_{0}  + 2\Delta s}
\\
b_{4} &amp; = &amp; b_{3}\pars{1 + r} + \pars{s_{0}  + 3\Delta s}
\\
\vdots &amp; = &amp; \vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\vdots
\end{array}
$$


In general we have to solve:
$$
b_{n} = b_{n - 1}\pars{1 + r} + \bracks{s_{0} + \pars{n - 1}\Delta s}\,,\quad n=2,3,4,\ldots\,;\qquad b_{1} = s_{0}\tag{1}
$$


  Lets $\quad\ds{{\rm B}\pars{z} \equiv \sum_{n = 1}^{\infty}b_{n}z^{n}}\quad$ with
  $\quad\ds{\verts{z} &lt; {1 \over 1 + r}}$:


\begin{align}
\sum_{n = 2}^{\infty}b_{n}z^{n} &amp;= \pars{1 + r}
\sum_{n = 2}^{\infty}b_{n - 1}z^{n}
+s_{0}\sum_{n = 2}^{\infty}z^{n} + \Delta s\sum_{n = 2}\pars{n - 1}z^{n}
\\[3mm]{\rm B}\pars{z} - b_{1}z &amp;= \pars{1 + r}\
\underbrace{\sum_{n = 1}^{\infty}b_{n}z^{n + 1}}_{\ds{=\ z\,{\rm B}\pars{z}}} +
s_{0}\,{z^{2} \over 1 - z} + \Delta s\,{z^{2} \over \pars{1 - z}^{2}}
\\[5mm]
\bracks{1 - \pars{r + 1}z}{\rm B}\pars{z}&amp;
=s_{0}\,{z \over 1 - z} + \Delta s\,{z^{2} \over \pars{1 - z}^{2}}
\end{align}


  $$
{\rm B}\pars{z}
={s_{0}\ z/\pars{1 - z} + \Delta s\ z^{2}/\pars{1 - z}^{2} \over 1 - \pars{r + 1}z}
$$
  $$
b_{n}=
\frac{\left[\left(r + 1\right)^{n} - n r-1\right]\Delta s  +
\left[\left(r + 1\right)^{n} - 1\right] r\,s_{0}}
{r^{2}}
$$
  $$
\color{#66f}{\large b_{n}}
=\color{#66f}{\large 12000\braces{1220\bracks{\pars{1201 \over 1200}^{n} - 1} - n}}
$$


$$
b_{284} \approx 499,325.84\,,\qquad
b_{\color{#c00000}{\Large 285}} \approx 502,781.94\,,\qquad
b_{286} \approx 506,250.93  
$$


  $$
\color{#c00000}{\Large 285} = \color{#66f}{\Large 5} \times 48
+ \color{#66f}{\Large 11} \times 4 + \color{#66f}{\Large 1}
$$


$$\color{#66f}{\large%
5\ \mbox{years}, 11\ \mbox{months and $1$ week}.  }
$$
Solution as an image:



Final Answer: