Differential equation is a type of equation involving of differentials.
More precisely, it involves of a function 
#card=math&code=y%20%3D%20f%28x%29&id=APCGl) and one or more of its derviatives (also possibly as well as the variable 
and some constant).
Examples of equations: (1) 
. Find x. (2) 
Example of differential equations. 
, here 
#card=math&code=y%3Dy%28x%29&id=lDBgj) and hence 
#card=math&code=%5Cfrac%7Bd%20y%7D%7Bd%20x%7D%3Dy%27%28x%29&id=hRUKd). To solve this differential eq. it is equivalent to finding a function 
#card=math&code=y%28x%29&id=pD5J0) such that 
%3Dx#card=math&code=y%27%28x%29%3Dx&id=XtCwE).
This is the problem of finding the anti-derivative of 
. All anti-derivatives of 
can be expressed by the indefinite integral 
%2BC#card=math&code=%5Cint%20x%20%5C%2C%20dx%20%3D%20%28x%5E2%2F2%29%2BC&id=qIPXM).
If we require in addition that 
%3D0#card=math&code=y%280%29%3D0&id=vCYuz) (for example), then from %3D(x%5E2%2F2)%2BC#card=math&code=y%28x%29%3D%28x%5E2%2F2%29%2BC&id=QR4QF), inserting 
, we get that %3D0%20%3D%20C#card=math&code=y%280%29%3D0%20%3D%20C&id=Mhian).
Another example: 
. %20%3D%20e%5Ex#card=math&code=y%28x%29%20%3D%20e%5Ex&id=kAXCA) is a solution to 
.
Abbrevation: ODE for ordinary differential equation. Here ordinary differential is to contrast to the “partial differential” in calculus of multivariable functions.
Initial Value Problem (IVP) is an ordinary differential equation plus an extra condition usually in the form of initial values, e.g. 
%20%3D%20y_0#card=math&code=y%28x_0%29%20%3D%20y_0&id=CGn0h).
%7D)%3D0#card=math&code=G%28x%2C%20y%2C%20y%5E%7B%28k%29%7D%29%3D0&id=Tjq2O). 
.
First order ODE 
%20%3D%20g(x)h(y)#card=math&code=y%27%20%3D%20f%28x%2C%20y%29%20%3D%20g%28x%29h%28y%29&id=LVAVT). <— we do not need to restrict ourselves that 
must be a funciton in 
. Rather, we just regard both 
and 
as symbols (or variables).
h(y)#card=math&code=y%27%20%3D%20%5Cfrac%7Bd%20y%7D%7Bd%20x%7D%20%3D%20g%28x%29h%28y%29&id=b71QR). 
%7D%20%3D%20%5Cint%20g(x)%5C%2C%20dx#card=math&code=%5Cint%20%5Cfrac%7Bd%20y%7D%7Bh%28y%29%7D%20%3D%20%5Cint%20g%28x%29%5C%2C%20dx&id=AG0Zx) 
%20%3D%20G(x%2C%20C)#card=math&code=H%28y%29%20%3D%20G%28x%2C%20C%29&id=Xt5q5)
If it is an initial value problem, then using the initial value to determine the constant 
.
Remark: In equations such as 
%20%3D%20x#card=math&code=e%5Ey%20%2B%20%5Csin%28y%29%20%3D%20x&id=o9PNa) , we cannot solve out 
explicitly. So we just write this expression as a solution to the ODE.
Remark: The trick to determine the constant in the general solution:
%20%5Cexp(4x%5E2%2B12x)#card=math&code=y-2%20%3D%20C_2%20%28y%2B2%29%20%5Cexp%284x%5E2%2B12x%29&id=tOguj). Using %3D-1#card=math&code=y%280%29%3D-1&id=W6U56) one deduces that 
and hence 
.
