First Order Differential Equations
First order differential equations are equations which include a first derivative ie) dy/dx if the equation is in terms of x and y.
Solving these equations involves finding a expression for one of the variables in terms of the other without including a derivative.
To do this usually a general solution is found, one which includes a constant from integration, and then using a given set of conditions (usually initial conditions) the particular solution can be found.
This is done by “splitting the variables” and then integrating the resulting equation. To do this you treat the derivative like a fraction and then multiply through by the bottom of that one derivative is on one side (eg dx) and the derivative is on the other side (eg dy), you then want all the x’s to be multiplied by the dx and all the y’s to multiplied by the dy such that there are no terms not being multiplied by either. You can then integrate both sides, the one by dx and the other by dy – remembering to add a constant to one side.
Then simply re arranging will give the general solution. If you need the particular solution substitute the given values into the equation to find out the value of the constant and then substitute this value for the constant into the general solution.
If all of that was a bit hard to follow here is s a worked example.
Example
Find the general solution to
dy/dx = 3x + 4
and the particular solution in the case y = 4 when x = 0 .
So we can rearrange to get
dy = (3x+4)dx
And then by integrating both sides we get
∫dy = ∫3x+4 dx
y = (3x+4)2/6 + c
Which is the general solution.
We can use boundary conditions, ie) y = 4 when x = 0 to find a particular solution
4 = (0+4)2/6 +c
c = 4-16/6 = 8/6 = 4/3
so the particular solution is
y = (3x+4)2/6 +4/3
Not sure that this is true:), but thanks for a post.
Joker
Oh yeah, on the above Ldi/dt + Ri=V, the quantity i is electric current flowing inside the circuit, L is self inductance of inductor, C is the capasitance of capasitor, R is resistance of the resistor, while V is the voltage used on charging process of the energy.
I am pleased http://rohedi.com is linked if you regard Stable Modulation Technique (SMT) and Bernoulli as alternative way of solving first order differential equation such as Bernoulli, arctan, and Chini differential equations are suitable to be promoted via this math blog. Thank you for this good appreciation. Best regards. Rohedi.
According to its general definition that Rohedi read at Mathematics Handbooks, First Order Differential Equation contains dependent function y and dy/dt of its derivative respect to the independent variable. Your example more related to integral case. Maybe appropiate example is the following first order linear differential equation Ldi/dt + Ri =V that is the charging model of energy into inductor using capasitor for the certain voltage. You can solve the DE’s example using Stable Modulation Technique (SMT) that postet at http://rohedi.com/content/view/25/26/. If all of you here need Bernoulli Integral for solving first order ODE, please click Rohedi’s address.
Thanks for you detailed comments,
I choose that example simply because it is typical of the questions in my A level maths exams.
Would it be ok to add a link to your site http://rohedi.com/
in the list called ‘maths sites’ on this blog,
Thanks david