There are several techniques for solving firstorder, linear differential equations. We list them here with links to other pages that discuss those techniques. But before you move on, let's discuss what a firstorder, linear ODE is and look at some easier techniques that will save you some time and energy.
Type 
Form 
Technique 

Homogeneous 
\(y' = f(y/x)\) 

Basic FirstOrder, Linear 
\( y' + p(t)y = g(t) \) 
What Is A FirstOrder, Linear Differential Equation?
A firstorder equation is one in which the highest derivative is a first derivative. A linear equation is a bit more complicated to determine. This short video clip explains this and shows some examples.
video by Dr Chris Tisdell 

How To Solve FirstOrder, Linear Differential Equations
Okay, now that you know what a firstorder, linear ODE is, we can tell you that there are several techniques to solve them, some of which are more rigorous than others. We listed a few at the top of the page but, before you jump there, take a minute to look at your differential equation. It may already be in a form that can be evaluated directly. If you can learn to recognize one of these forms, it may save you some time and energy on your homework or exam. Also, it will help you to start to get a feel for what to look for.
Sometimes a differential equation may already be in either derivative product rule or quotient rule form. This video clip shows some examples and explains what to look for.
video by Dr Chris Tisdell 

Here are a couple of videos containing indepth discussions of firstorder, linear differential equations with constant coefficients. Given an equation of the form \(y' + ky = g(t)\), the solution is \(\displaystyle{ y = e^{kt}\int{e^{kt}g(t)~dt} }\). This is easily found from the use of integrating factors. The videos have discussion relating to steadystate and transient solutions and some specific examples of functions for \(g(t)\). They are good to watch to get a feel for these types of differential equations.
video by MIT OCW 

video by MIT OCW 

How To Solve Any FirstOrder Differential Equation
This is a good video to watch to get an idea of what is coming up in your study of firstorder equations. It puts together the techniques in an entertaining and informative way. However, you will not understand most of it until you have studied more differential equations techniques. Once you have more experience with firstorder differential equations, come back here and watch this video again.
video by Dr Chris Tisdell 

Okay, so you are now ready to start learning specific techniques for solving firstorder, linear differential equations. We suggest substitution or integrating factors.
You CAN Ace Differential Equations
external links you may find helpful 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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