Answer:
Following are the solution to the given points:
Step-by-step explanation:
p= amount at any time
then [tex]\frac{dp}{dt}=0.07 p[/tex]
[tex]p=p_0 (1+e^{0.07 t) \\\\[/tex]
[tex]P_0=\ principle \ amount \\\\ interest \ =p-p_0=p_0e^{0.07t}\\\\ p=y\frac{dy}{dt}=0.07y_0[/tex]
at time [tex]= t\ years[/tex]
[tex]\frac{dp}{dt}\\\\c_0=\$ \ 20000\ year\\\\p \to p+dp\\\\ dp=0.07p+c_0[/tex]
y= amount at any times
[tex]\frac{dy}{dt}=0.07y+c_0\\\\dt= \frac{dy}{c_0+0.07y}[/tex]
For point a:
[tex]\frac{dy}{dt}=20000+0.07 y[/tex]
For point b:
[tex]\frac{dy}{20000+0.07 y}=dt\\\\\frac{1}{0.07}|n(\frac{0.07y+20000}{20000})=t\\\\\frac{0.07y}{20000}+1=e^{0.07t}\\\\y=7.1428\times 10^4[e^{0.07t} -1]\\\\y=\frac{20000}{0.07}[e^{0.07t}-1]\\\\[/tex]
For point c:
[tex]y(30)=827391.75[/tex]
