Answer:
Part a)
[tex]y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}[/tex]
Part b)
Check the attached figure to see the ultimate behavior of the graph.
Part c)
The level = 25, Amplitude = 0.2499
Step-by-step Solution:
Part a)
Given:
[tex]Q(0)=50[/tex]
Rate in:
[tex]\frac{1}{4}\left(1+\frac{1}{2}sint\right)\cdot 2\:=\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)[/tex]
Rate out:
[tex]\frac{Q}{100}\cdot 2=\frac{Q}{50}[/tex]
So, the differential equation would become:
[tex]\frac{dQ}{dt}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)-\frac{Q}{50}[/tex]
Rewriting the equation:
[tex]\frac{dQ}{dt}+\frac{Q}{50}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)[/tex]
As [tex]p(x)[/tex] is the coefficient of [tex]y[/tex], while [tex]q(x)[/tex] is the constant term in the right side of the equation:
[tex]p\left(x\right)=\frac{1}{50}[/tex]
[tex]q\left(x\right)=\frac{1}{2}\left(1+\frac{1}{2}sint\right)[/tex]
First it is important to determine the function [tex]\mu[/tex] :
[tex]\mu \left(t\right)=e^{\int \:p\left(t\right)dt}[/tex]
[tex]=e^{\int \:\left(\frac{1}{50}\right)dt}[/tex]
[tex]=e^{\frac{t}{50}}[/tex]
The general solution then would become:
[tex]y\left(t\right)=\frac{1}{\mu \left(t\right)}\left(\int \mu \left(t\right)q\left(t\right)dt+c\:\right)[/tex]
[tex]=\frac{1}{e^{\frac{t}{50}}}\int e^{\frac{t}{50}}\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)dt+\frac{1}{e^{\frac{t}{50}}}c[/tex]
[tex]=\frac{1}{e^{\frac{t}{50}}}\left(\frac{-25e^{\frac{t}{50}}\left(50cost-sint\right)}{5002}+25e^{\frac{t}{50}}\right)+\frac{1}{e^{\frac{t}{50}}}c[/tex]
[tex]=\frac{\left-1250cost+25sint\right}{5002}+25+\frac{1}{e^{\frac{t}{50}}}c[/tex]
Evaluate at [tex]t=0[/tex]
[tex]50=y\left(0\right)=\frac{\left(-1250cos0+25sin0\right)}{5002}+25+\frac{1}{e^{\frac{0}{50}}}c[/tex]
Solve to c:
[tex]c=25+\frac{1250}{5002}[/tex]
[tex]\mathrm{Cancel\:}\frac{1250}{5002}:\quad \frac{625}{2501}[/tex]
[tex]c=25+\frac{625}{2501}[/tex]
[tex]\mathrm{Convert\:element\:to\:fraction}:\quad \:25=\frac{25\cdot \:2501}{2501}[/tex]
[tex]c=\frac{25\cdot \:2501}{2501}+\frac{625}{2501}[/tex]
[tex]\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}[/tex]
[tex]c=\frac{25\cdot \:2501+625}{2501}[/tex]
[tex]c=\frac{63150}{2501}[/tex]
[tex]c\approx 25.25[/tex]
Therefore, the general solution then would become:
[tex]y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}[/tex]
Part b) Plot the Solution to see the ultimate behavior of the graph
The graph appears to level off at about the value of [tex]Q=25[/tex].
The graph is attached below.
Part c)
In the graph we note that the level is [tex]Q=25[/tex].
Therefore, the level = 25
The amplitude is the (absolute value of the) coefficient of [tex]cost\:t[/tex] in the general solution (as the coefficient of the sine part is a lot smaller):
Therefore,
[tex]A=\frac{1250}{5002}\:\approx 2.499[/tex]
Keywords: differential equation, word problem
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