This week we covered separation of variables. The idea of this concept was to basically get the y's on one side and the x's on one side, and then anti-deriv both sides. An example of this would be dy/dx=(x^2)/(1-y^2) find x such that y(0)=1. First I'd multiply both sides by (1-y^2) and dx to get them out of the denomonater and on the other side. My equation would be (1-y^2)dy=(x^2)dx. Then I'd anti-deriv both sides to get y-(y^3/3)=(x^3/3)+c. Next, I'd plug in the point given (0,1) to get (1)-((1)^3/3)=(0^3/3)+c. Then I'd solve for c to get 2/3. Finally, I would rewrite the anti-derived equation plus the c we just found to get y-(y^3/3)=(x^3/3)+2/3. You could rewrite this to make it look better by multiplying it all by 3 to cancel out the 3 as a denomonater. My final answer would be 3y-y^3=x^3+2.
The second part of section 6.4 was on exponential growth and decay. We were given the following equations:
dy/dt=k*y y=yo*e^kt y=yo*e^-kt
These types of problems are pretty easy too. You'd use the first one if they gae you k, t, and y. You'd use the 2nd one for exponential growth, and the last one for exponential decay. An example using the first equatioin would be if you were given that k=1.5 when t=0 and y=100. First you would plug in 1.5 for k to get dy/dt=1.5y. Then you would use separation of variables to get the y's on one side and the x's on the other, so you'd get (1/y)dy=1.5dt. Then take the anti-derivative to get ln(y)=1.5t+c. Then plug in the given t and y values to get ln(100)=1.5(0)+c. Solve for c to get that c=ln(100). Then rewrite the anti-derived equation with the c we solved for to get ln(y)=1.5t+ln(100). You could leave the answer as that, or you could rewrite it to look nice to be y=100e^1.5t.
The second part of section 6.4 was on exponential growth and decay. We were given the following equations:
dy/dt=k*y y=yo*e^kt y=yo*e^-kt
These types of problems are pretty easy too. You'd use the first one if they gae you k, t, and y. You'd use the 2nd one for exponential growth, and the last one for exponential decay. An example using the first equatioin would be if you were given that k=1.5 when t=0 and y=100. First you would plug in 1.5 for k to get dy/dt=1.5y. Then you would use separation of variables to get the y's on one side and the x's on the other, so you'd get (1/y)dy=1.5dt. Then take the anti-derivative to get ln(y)=1.5t+c. Then plug in the given t and y values to get ln(100)=1.5(0)+c. Solve for c to get that c=ln(100). Then rewrite the anti-derived equation with the c we solved for to get ln(y)=1.5t+ln(100). You could leave the answer as that, or you could rewrite it to look nice to be y=100e^1.5t.