On March 11 we studied differential equations. These are defined as any equation involving one or more derivatives of an unknown curve/function. There are simple, medium, and tough differential equations to find. The simplest being something we already know how to do. If we are given dy/dx = 2x, the differential equation is just the anti-derivative of this so the general solution (using variables) would be y=x^2 + c or the particular solution (one where you have a number in for the previously unknown constants) would be y=x^2 + 4 if first given an initial condition such as y(1) = 5. A tougher one to solve would be if given dy/dx = -x/4y and the initial condition is that one of the points on the curve is (0,-3). To solve this we’ll have to separate variables so the y’s are on one side and the x’s are on the other. So to solve this you’d cross multiply to get dy4y = dx(-x). Next we can put an integral sign in front of both parts of the equation, so we can solve for the anti-derivative. So eventually we receive 2y^2 = -.5x^2 + C (general solution). Then we solve using the given point to find that C= 18 making the particular solution 2y^2 = .5x^2 + 18. Another example for you to solve using a math application question would be the modeling of a biological population. If you are given dP/dt = kP, P(0) = 400, and P(3) = 1200 and you already know that P is the number of yeast cells in a population, k is an unknown constant, and p must be greater than 0, you may solve. The answer can be written either P(t) = 400 (e^ ln 3) ^ t/3 or P(t) = 400 (3)^t/3 and they both mean the same thing.
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On March 11 we studied differential equations. These are defined as any equation involving one or more derivatives of an unknown curve/function. There are simple, medium, and tough differential equations to find. The simplest being something we already know how to do. If we are given dy/dx = 2x, the differential equation is just the anti-derivative of this so the general solution (using variables) would be y=x^2 + c or the particular solution (one where you have a number in for the previously unknown constants) would be y=x^2 + 4 if first given an initial condition such as y(1) = 5. A tougher one to solve would be if given dy/dx = -x/4y and the initial condition is that one of the points on the curve is (0,-3). To solve this we’ll have to separate variables so the y’s are on one side and the x’s are on the other. So to solve this you’d cross multiply to get dy4y = dx(-x). Next we can put an integral sign in front of both parts of the equation, so we can solve for the anti-derivative. So eventually we receive 2y^2 = -.5x^2 + C (general solution). Then we solve using the given point to find that C= 18 making the particular solution 2y^2 = .5x^2 + 18. Another example for you to solve using a math application question would be the modeling of a biological population. If you are given dP/dt = kP, P(0) = 400, and P(3) = 1200 and you already know that P is the number of yeast cells in a population, k is an unknown constant, and p must be greater than 0, you may solve. The answer can be written either P(t) = 400 (e^ ln 3) ^ t/3 or P(t) = 400 (3)^t/3 and they both mean the same thing.
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