FirstOrder Differential Equations
Direct Integration (linear ODEs only)
Unfortunately, direct integration techniques can only be used on ODE's of the form
Unfortunately, direct integration techniques can only be used on ODE's of the form
\(\frac{dy}{dx} = f(x).\)

(1)

Since this is a course on differential equations, and you have already dealt with equation like (1) in calculus, you can expect to not see this method on a final exam in this course.
Separation of Variables
This method can be used anytime you have an ODE that can be written in the form
This method can be used anytime you have an ODE that can be written in the form
\(\frac{dy}{dx} = f(x) g(y).\)

(2)

However, it is desirable that \(f(x)g(y)\) and \([f(x)g(y)]_y\) be continuous as this guarantees a unique solution to an IVP. Note that \(g(y)\) in (2) need not be linear! Thus, separation of variables can be used on nonlinear ODEs.
Integrating Factors (linear ODEs only)
This is the method of preference when the given ODE can be written in the standard form
This is the method of preference when the given ODE can be written in the standard form
\(\frac{dy}{dx} + p(x) y = q(x),\)

(3)

where we again have the desired restriction that \(p(x)y + q(x)\) and \([p(x) y + q(x)]_y\) are continuous so we may guarantee a unique solution to a given IVP.
Without rederiving the entire method, the use of an integrating factor stems from noticing that the ODE (3) is very close in form to \(\frac{d}{dx}\left(\mu (x) y\right) = f(x)\). If we imagine multiplying both sides of (3) by a fortuitous choice for \(\mu (x)\), we could arrive at \[\frac{d}{dx}\left(\mu (x) y\right) = \mu (x) q(x).\]This ODE is directly integrable. Hence, the solution from here is trivial. It has already been shown that our choice for the integrating factor is
Without rederiving the entire method, the use of an integrating factor stems from noticing that the ODE (3) is very close in form to \(\frac{d}{dx}\left(\mu (x) y\right) = f(x)\). If we imagine multiplying both sides of (3) by a fortuitous choice for \(\mu (x)\), we could arrive at \[\frac{d}{dx}\left(\mu (x) y\right) = \mu (x) q(x).\]This ODE is directly integrable. Hence, the solution from here is trivial. It has already been shown that our choice for the integrating factor is
\(\mu (x) = e^{\int{p(x) dx}}.\)

(4)

WARNING! It is important to note that the original ODE must be rewritten in the standard form (3) prior to using this method.
Exact Equations (a.k.a. Integration of Partials)
This can be one of the easiest methods to use; however, it is also one of the most difficult to identify when it can be used. The most obvious ODE to try this method on is one that is already in the differential form
This can be one of the easiest methods to use; however, it is also one of the most difficult to identify when it can be used. The most obvious ODE to try this method on is one that is already in the differential form
\(X(x,y) dx + Y(x,y) dy = 0.\)

(5)

Identification of whether or not to use this method becomes a little shakier when the ODE is in the general form\[X(x,y) + Y(x,y) \frac{dy}{dx} = 0,\]or the standard form\[\frac{dy}{dx} + R(x,y) = 0.\]The use of the method requires the ODE to be exact. That is, we require
\(X_y = Y_x.\)

(6)

If this can be guaranteed, then we can guarantee that \(X(x,y) = f_x\) and \(Y(x,y) = f_y\). Solving the ODE is as simple as integrating either \(X(x,y)\) with respect to \(x\) or \(Y(x,y)\) with respect to \(y\) (while holding the other variable constant). For ease of discussion, let's just say we are integrating \(X(x,y)\) with respect to \(x\). Then we would get \(f(x,y) = \bar{X} + g(y)\), where \(\bar{X}\) is the antiderivative of \(X\) (with respect to \(x\)) and \(g(y)\) is playing the role of the constant of integration. We then find \(f_y\) through simple partial differentiation and compare terms to see what \(g'(y)\) is. At this point, we integrate \(g'(y)\) with respect to \(y\) to finally get that missing \(g(y)\). We now have our final equation/solution which is \(f(x,y) = C\).
This method sounds harder than it actually is. It's worth doing several examples to get the hang of it. Once you get it, it is a super simple method (albeit, not always usable).
If it happens that \(X_y \neq Y_x\), then we have some work ahead of us. I would try another method (if one is available). Otherwise, you can sometimes force an equation to become exact by using an integrating factor. If we multiply both sides of (5) by a yet to be determined function \(\mu\), then (5) becomes\[\left(\mu X\right) dx + \left(\mu Y\right) dy = 0.\]For this to be exact, we need\[\left(\mu X\right)_y = \left(\mu Y\right)_x,\]which can be written (using the Product Rule) as
This method sounds harder than it actually is. It's worth doing several examples to get the hang of it. Once you get it, it is a super simple method (albeit, not always usable).
If it happens that \(X_y \neq Y_x\), then we have some work ahead of us. I would try another method (if one is available). Otherwise, you can sometimes force an equation to become exact by using an integrating factor. If we multiply both sides of (5) by a yet to be determined function \(\mu\), then (5) becomes\[\left(\mu X\right) dx + \left(\mu Y\right) dy = 0.\]For this to be exact, we need\[\left(\mu X\right)_y = \left(\mu Y\right)_x,\]which can be written (using the Product Rule) as
\(\mu_y X + \mu X_y = \mu_x Y + \mu Y_x.\)

(7)

At this point, I will just say that we require \(\mu\) to either be a function of \(x\) or a function of \(y\), but not both. For simplicity of the remainder of this discussion, suppose that \(\mu = \mu (x)\). Thus, the partial \(\mu_y\) in (7) will disappear. Solve the resulting equation for \(\mu_x\). If the resulting equation can be written solely in terms of \(x\), then we can find \(\mu\) via separation of variables. We then multiply both sides of (5) by this newly found \(\mu\) and solve using Integration of Partials. Yes... it's longwinded.