Respuesta :
Answer: The required solution is [tex]y=50e^{0.1386t}.[/tex]
Step-by-step explanation:
We are given to solve the following differential equation :
[tex]\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.
From equation (i), we have
[tex]\dfrac{dy}{y}=kdt.[/tex]
Integrating both sides, we get
[tex]\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}][/tex]
Also, the conditions are
[tex]y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50[/tex]
and
[tex]y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.[/tex]
Thus, the required solution is [tex]y=50e^{0.1386t}.[/tex]
Answer:
[tex]\frac{1}{5}\ln(2)=k[/tex]
Solution without isolating [tex]y[/tex]:
[tex]\ln|y|=\frac{1}{5}\ln(2)x+\ln(50)[/tex]
Solution with isolating [tex]y[/tex]:
[tex]y=50 \cdot 2^{\frac{1}{5}x}[/tex]
Step-by-step explanation:
[tex]\frac{dy}{dx}=ky[/tex]
We will separate the variables so we can integrate both sides.
Multiply [tex]dx[/tex] on both sides:
[tex]dy=ky dx[/tex]
Divide both sides by [tex]y[/tex]:
[tex]\frac{dy}{y}=k dx[/tex]
Now we may integrate both sides:
[tex]\ln|y|=kx+C[/tex]
The first condition says [tex]y(0)=50[/tex].
Using this into our equation gives us:
[tex]\ln|50|=k(0)+C[/tex]
[tex]\ln|50|=C[/tex]
So now our equation is:
[tex]\ln|y|=kx+\ln(50)[/tex]
The second condition says [tex]y(5)=100[/tex].
Using this into our equation gives us:
[tex]\ln|100|=k(5)+\ln(50)[/tex]
[tex]\ln(100)=k(5)+\ln(50)[/tex]
Let's find [tex]k[/tex].
Subtract [tex]\ln(50)[/tex] on both sides:
[tex]\ln(100)-\ln(50)=k(5)[/tex]
I'm going to rewrite the left hand side using quotient rule for logarithms:
[tex]\ln(\frac{100}{50})=k(5)[/tex]
Reducing fraction:
[tex]\ln(2)=k(5)[/tex]
Divide both sides by 5:
[tex]\frac{\ln(2)}{5}=k[/tex]
[tex]\frac{1}{5}\ln(2)=k[/tex]
So the solution to the differential equation satisfying the give conditions is:
[tex]\ln|y|=\frac{1}{5}\ln(2)x+\ln(50)[/tex]
Most likely they will prefer the equation where [tex]y[/tex] is isolated.
Let's write our equation in equivalent logarithm form:
[tex]y=e^{\frac{1}{5}\ln(2)x+\ln(50)}[/tex]
We could rewrite this a bit more.
By power rule for logarithms:
[tex]y=e^{\ln(2^{\frac{1}{5}x})+\ln(50)}[/tex]
By product rule for logarithms:
[tex]y=e^{\ln(2^{\frac{1}{5}x} \cdot 50)}[/tex]
Since the natual logarithm and given exponential function are inverses:
[tex]y=2^{\frac{1}{5}x} \cdot 50[/tex]
By commutative property of multiplication:
[tex]y=50 \cdot 2^{\frac{1}{5}x}[/tex]
