Respuesta :
Answer:
The population size after four hours is [tex]N\left(4\right)=1.4800[/tex] millions of individuals per mL.
Step-by-step explanation:
Because [tex]0.57=\frac{57}{100}[/tex] and [tex]0.0026=\frac{13}{5000}[/tex] we can say that the differential equation is [tex]\frac{dN}{dt} =(\frac{57}{100}-\frac{13}{5000}y )y[/tex].
To find [tex]N(4)[/tex] for [tex]\frac{dN}{dt} =(\frac{57}{100}-\frac{13}{5000}y )y[/tex], when [tex]N(0) = 0.2=\frac{1}{5}[/tex] with a step size of [tex]h = 0.5 =\frac{1}{2}[/tex] using the Euler's method you must:
The Euler's method states that [tex]y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)[/tex], where [tex]t_{n+1}=t_n + h[/tex].
We have that [tex]h=\frac{1}{2}[/tex], [tex]t_0=0[/tex], [tex]f(t,y)=y \left(\frac{57}{100} - \frac{13 y}{5000}\right)[/tex].
Step 1.
[tex]t_{1}=t_{0}+h=0+\frac{1}{2}=\frac{1}{2}[/tex]
[tex]y\left(t_{1}\right)=y\left( \frac{1}{2} \right)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=\frac{1}{5}+h \cdot f \left(0, \frac{1}{5} \right)=\frac{1}{5} + \frac{1}{2} \cdot \left(0.113896 \right)=0.25695[/tex]
Step 2.
[tex]t_{2}=t_{1}+h=\frac{1}{2}+\frac{1}{2}=1[/tex]
[tex]y\left(t_{2}\right)=y\left( 1 \right)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=0.256948+h \cdot f \left(\frac{1}{2}, 0.256948 \right)=0.33009[/tex]
Step 3.
[tex]t_{3}=t_{2}+h=1+\frac{1}{2}=\frac{3}{2}[/tex]
[tex]y\left(t_{3}\right)=y\left( \frac{3}{2} \right)=y_{3}=y_{2}+h \cdot f \left(t_{2}, y_{2} \right)=0.33009+h \cdot f \left(1, 0.33009 \right)=0.42403[/tex]
Step 4.
[tex]t_{4}=t_{3}+h=\frac{3}{2}+\frac{1}{2}=2[/tex]
[tex]y\left(t_{4}\right)=y\left( 2 \right)=y_{4}=y_{3}+h \cdot f \left(t_{3}, y_{3} \right)=0.42403+h \cdot f \left(\frac{3}{2}, 0.42403 \right)=0.54464[/tex]
Step 5.
[tex]t_{5}=t_{4}+h=2+\frac{1}{2}=\frac{5}{2}[/tex]
[tex]y\left(t_{5}\right)=y\left( \frac{5}{2} \right)=y_{5}=y_{4}+h \cdot f \left(t_{4}, y_{4} \right)=0.54464+h \cdot f \left(2, 0.54464 \right)=0.69948[/tex]
Step 6.
[tex]t_{6}=t_{5}+h=\frac{5}{2}+\frac{1}{2}=3[/tex]
[tex]y\left(t_{6}\right)=y\left( 3 \right)=y_{6}=y_{5}+h \cdot f \left(t_{5}, y_{5} \right)=0.69948+h \cdot f \left(\frac{5}{2}, 0.69948 \right)=0.89819[/tex]
Step 7.
[tex]t_{7}=t_{6}+h=3+\frac{1}{2}=\frac{7}{2}[/tex]
[tex]y\left(t_{7}\right)=y\left( \frac{7}{2} \right)=y_{7}=y_{6}+h \cdot f \left(t_{6}, y_{6} \right)=0.89819+h \cdot f \left(3, 0.89819 \right)=1.153129[/tex]
Step 8.
[tex]t_{8}=t_{7}+h=\frac{7}{2}+\frac{1}{2}=4[/tex]
[tex]y\left(t_{8}\right)=y\left( 4 \right)=y_{8}=y_{7}+h \cdot f \left(t_{7}, y_{7} \right)=1.153129+h \cdot f \left(\frac{7}{2}, 1.153129 \right)=1.4800[/tex]
Therefore, the population size after four hours is [tex]N\left(4\right)=1.4800[/tex] millions of individuals per mL.
