Solving Diff. Equations: x(t) and y(t)

[circa415]

Given x(0)=-2, y(0)=1, x'(t)=2x(t), y'(t)=-2y(t), solve for x(t) and y(t)

I plugged this into mathematica and got -2e^(2t)=x(t) and e^(-2t) = y(t)

I notice that x'(t) = 2x(t) means that x(t) has to be e^t and there must be a coefficient (2) in front of the t because the only derivative that has this property is e^x. And the coefficient in front of e will be given by x(0). But is there a "method" that I can go about doing this? or can you only do it by observation?

 

[stunner5000pt]

for the x(t) you forgot to include the x(0) part in the formation of the x(t) function, you need to add something to it to be true for x(0)

Similarly for the y(0)

 

[wais]

There are multiple methods for solving differential equations, each with its own advantages and limitations. In this case, the equations are separable, which means we can separate the variables and integrate each side separately.

Starting with x'(t) = 2x(t), we can divide both sides by x(t) and integrate from 0 to t:

1/x(t) dx = 2 dt

Integrating both sides gives us:

ln|x(t)| = 2t + C

Using the initial condition x(0) = -2, we can solve for the constant C:

ln|-2| = 2(0) + C

C = ln(2)

Therefore, our solution for x(t) is:

x(t) = e^(2t + ln(2))

Simplifying this, we get:

x(t) = 2e^(2t)

Similarly, for y'(t) = -2y(t), we can divide both sides by y(t) and integrate from 0 to t:

1/y(t) dy = -2 dt

Integrating both sides and using the initial condition y(0) = 1, we get:

ln|y(t)| = -2t + C

Using the initial condition, we can solve for the constant C:

ln|1| = -2(0) + C

C = 0

Therefore, our solution for y(t) is:

y(t) = e^(-2t)

Combining these solutions with the given initial conditions, we get:

x(t) = -2e^(2t)

y(t) = e^(-2t)

So, using the method of separation of variables, we were able to solve for x(t) and y(t) in terms of t. Other methods such as the method of undetermined coefficients or variation of parameters can also be used to solve these types of differential equations. It is important to familiarize yourself with multiple methods and choose the most appropriate one for each specific problem.