Method: Separable Differential Equations Exponential Models

There is no difference between this method and the method you learned in the previous Separable Differential Equations. Really the only reason that they break this up into two different sections is that these problems require the application of some Laws of Logarithms and some Laws of Exponents in order to get your final result.

• Step 1: Identify your two variables.
• Step 2: Separate your equation by variables on to either side of the equals.
• This will mean treating your differential notation , $\frac{\mathit{dy}}{\mathit{dx}}$, like two variables being divided. You will would want the $\mathit{dx}$ piece to be on the side with the x’s , or the $\mathit{dy}$ piece to be on the side with the y’s , or the $\mathit{dt}$ to be on the side with the t’s . Make sure that you are always multiplying or dividing variables by the $\mathit{dx},\mathit{ dy},\mathit{ dt}$ piece. You are about to take an antiderivative, and an antiderivative is normally of the form $\int \mathit{equation\; dx}$. You are never adding or subtracting your $\mathit{dx},\mathit{ dy},\mathit{ dt}$ piece.
• Step 3: Take the antiderivative of both sides of the equation.

This antiderivative process will be the same as a standard indefinite integral process. You will only need a single +C value that will work for both antiderivatives. It is best to put this +C on the independent variable side ( x -variable) of the equation. Although it does not matter at this step of the process which side you choose.

This type of problem will usually result in an $e^x$ equation or an $\mathit{ln}\left(x\right)$antiderivative.

• Step 4: Solve the equation you found in Step 3 to get the dependent variable ( y -variable) alone on one side of the equals.

It is this step where you will run into the need to apply some Laws of Logarithms and some Laws of Exponents .

• Step 5 (If asked to find a particular solution): Sometimes you will be given an actual value or point that the solution needs to go through (i.e., an $(x,y)$). You will use that value to find the +C value for your specific situation. This is just like the method you would apply in an initial value problem you learned earlier.