y'(t) = f(t) 
(Required background: It is assumed that you are already familiar with derivatives and antiderivatives.)
What is a differential equation? A differential equation is an equation that involves at least one derivative. A differential equation is a mathematical model that describes how something is changing (or something's rate of change, hence it involves a derivative). Let's say you observe that the number of frogs in pond is increasing at a rate given by the equation 9t^{2}4t, where "t" is the number of weeks after the first day of spring. If we let y(t) be the number of frogs in the pond, this observation can be modeled by the following differential equation: y'(t) = 9t^{2}4t. Since this differential equation is also "just a derivative" we can solve it by doing an antiderivative. (In the next section you will see how differential equations are normally different from being "just a derivative".) Thus after "doing the antiderivative" we can see that y(t) = 3t^{3}2t^{2}+C. If we want to be more formal about our work we can also solve the differential equation by setting up an integral. The solution is still: y(t) = 3t^{3}2t^{2}+C. Let's say that you also know that there were 12 frogs initially, y(0) = 12, you can use this information to solve for "C". y(0) = 3(0)^{3}2(0)^{2}+C = 12 C = 12 y(t) = 3t^{3}2t^{2}+12. Now this solution can be used to predict future numbers of frogs in the pond. How many frog will there be 10 weeks after the first day of spring? y(10) = 3(10)^{3}2(10)^{2}+12 y(10) = 3000200+112 = 2812. This solution predicts that there will be 2812 frogs in the pond 10 weeks after the first day of spring. 
Tips on vocabulary The Differential Equation is the " red thing" above with the derivative in front of the equal sign. The General solution is the "green thing" with the "+C" at the end. It is called the general solution because in general you know what the formula or solution is, you just don't know what "C" is yet. The Initial condition is the "purple thing" above. The initial condition doesn't have to be given at time "zero", but it is given information that will help you find the constant from integrating, the "C". If the problem doesn't have an initial condition then you can only find the general solution. The Particular solution is the "blue thing" above. Once you determine the particular value of the constant, the "C", that matches the initial condition, the general solution becomes the particular solution. Rewrite the general solution replacing "C" with it's known value. The Answer is the 2812 frogs. Yes this is the "answer" but it is probably the least important step of the problem. How you get it and how much work you show is far more important than what you actually get. 
More adviceDon't Go In Public Areas! This advice applies to math too. In fact if you can remember "Don't Go In Public Areas", the first letter in each word is meant to help you remember the five steps to solving differential equations. 5 Step to Solving Differential Equations Step 1. Find the Differential Equation. Step 2. Write down the General Solution. Step 3. Look for Initial Conditions. Step 4. Get the Particular Solution (using the initial conditions). Step 5. Only after completing the above steps may you Answer the question. 
Sample Problem Let's try applying this to a typical problem. The acceleration of a rocket t seconds after launch is 12t4 m/s per second. If the initial speed was 3 m/s and 1 second after launch the rocket was at an elevation of 20 meters, what is the rocket's height 10 seconds after launch? Step 1. Find the Differential Equation. Since acceleration is the second derivative of height, we can write: h"(t) = 12t4, this is a second order differential equation. Step 2. Write down the General Solution. Since we have a second order differential equation in term of "t" we can solve this kind of differential equation by integrating twice. h'(t) = 6t^{2}4t+C, and h(t) = 2t^{3}2t^{2}+Ct+K Step 3. Look for Initial Conditions. We started with a second order differential equation and therefore have two constants from integrating. We need to find two initial conditions. Since speed is the first derivative of height, the initial speed being 3m/s translates into h'(0) = 3 m/s. The fact that the height was 20 meters after 1 second means h(1) = 20 meters. Step 4. Get the Particular Solution (using the initial conditions). Now that you have the initial conditions, use them to get the particular solution. To use the h'(0) = 3 you need to plug the "0" into the first derivative, h'(t) = 6t^{2}4t+C and set it equal to "3". This gives you "C". See below. h'(0) = 3 h'(t) = 6t^{2}4t+C h'(0) = 6(0)^{2}4(0)+C = 3 C = 3. Next use the other initial condition to solve for "K". Since this initial condition is not primed, it goes directly into the general solution. h(1) = 20 h(t) = 2t^{3}2t^{2}+Ct+K h(1) = 2(1)^{3}2(1)^{2}+3(1)+K = 20 22+3+K = 20 K = 17 This gives us h(t) = 2t^{3}2t^{2}+3t+17 as the particular solution. Step 5. Answer the question. What is the rocket's height 10 seconds after launch? To answer this, just plug "10" into the particular solution. h(10) = 2(10)^{3}2(10)^{2}+3(10)+17 = 2000200+30+17 = 1847 meters. 
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