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Sunday, December 29, 2013

Solving Equations - Homogeneous Functions, 2

Category: Differential Equations, Integral Calculus

"Published in Newark, California, USA"

Find the general solution for


Solution: 

Consider the given equation above


Did you notice that the given equation cannot be solved by separation of variables? The first term is a combination of x and y in the group and there's no way that we can separate x and y. 

This type of differential equation is a homogeneous function. Let's consider this procedure in solving the given equation as follows


Let


so that


Substitute the values of y and dy to the given equation, we have








The resulting equation can now be separated by separation of variables as follows



Integrate on both sides of the equation, we have




But



Hence, the above equation becomes






Therefore, the general solution is

 

Saturday, December 28, 2013

Variable Separation, 6

Category: Differential Equations, Integral Calculus

"Published in Newark, California, USA"

Find the general solution for 


Solution:

Consider the given equation above


Transpose xy to the right side of the equation, we have



Arrange the above equation by separation of variables, we have




Integrate on both sides of the equation, we have










Take the inverse natural logarithm on both sides of the equation




where K = eC. Therefore, the general solution is

 

Friday, December 27, 2013

Solving 3rd Order Differential Equations

Category: Differential Equations, Integral Calculus

"Published in Newark, California, USA"

Find the general solution for


Solution:

Consider the given equation above


The given equation is a 3rd Order Differential Equation because the third derivative of y with respect to x is involved. We can rewrite given equation as follows



Multiply both sides of the equation by dx, we have




Integrate on both sides of the equation, we have




Rewrite the above equation as follows



Multiply both sides of the equation by dx, we have




Integrate on both sides of the equation, we have




Multiply both sides of the equation by dx, we have




Integrate on both sides of the equation, we have





where B = ½ C1. Therefore, the general solution is