Question

Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) = 100

Answer 1

Answer:  The required solution is $y=50e^{0.1386t}.$

Step-by-step explanation:

We are given to solve the following differential equation :

$\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

$\dfrac{dy}{y}=kdt.$

Integrating both sides, we get

$\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]$

Also, the conditions are

$y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50$

and

$y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.$

Thus, the required solution is $y=50e^{0.1386t}.$

Answer 2

Answer:

$\frac{1}{5}\ln(2)=k$

Solution without isolating $y$:

$\ln|y|=\frac{1}{5}\ln(2)x+\ln(50)$

Solution with isolating $y$:

$y=50 \cdot 2^{\frac{1}{5}x}$

Step-by-step explanation:

$\frac{dy}{dx}=ky$

We will separate the variables so we can integrate both sides.

Multiply $dx$ on both sides:

$dy=ky dx$

Divide both sides by $y$:

$\frac{dy}{y}=k dx$

Now we may integrate both sides:

$\ln|y|=kx+C$

The first condition says $y(0)=50$.

Using this into our equation gives us:

$\ln|50|=k(0)+C$

$\ln|50|=C$

So now our equation is:

$\ln|y|=kx+\ln(50)$

The second condition says $y(5)=100$.

Using this into our equation gives us:

$\ln|100|=k(5)+\ln(50)$

$\ln(100)=k(5)+\ln(50)$

Let's find $k$.

Subtract $\ln(50)$ on both sides:

$\ln(100)-\ln(50)=k(5)$

I'm going to rewrite the left hand side using quotient rule for logarithms:

$\ln(\frac{100}{50})=k(5)$

Reducing fraction:

$\ln(2)=k(5)$

Divide both sides by 5:

$\frac{\ln(2)}{5}=k$

$\frac{1}{5}\ln(2)=k$

So the solution to the differential equation satisfying the give conditions is:

$\ln|y|=\frac{1}{5}\ln(2)x+\ln(50)$

Most likely they will prefer the equation where $y$ is isolated.

Let's write our equation in equivalent logarithm form:

$y=e^{\frac{1}{5}\ln(2)x+\ln(50)}$

We could rewrite this a bit more.

By power rule for logarithms:

$y=e^{\ln(2^{\frac{1}{5}x})+\ln(50)}$

By product rule for logarithms:

$y=e^{\ln(2^{\frac{1}{5}x} \cdot 50)}$

Since the natual logarithm and given exponential function are inverses:

$y=2^{\frac{1}{5}x} \cdot 50$

By commutative property of multiplication:

$y=50 \cdot 2^{\frac{1}{5}x}$