You will, on occasion, be presented with differential equations that have linear coefficients. This means that the coefficients themselves describe algebraic linear equations. When equated to zero they will either define parallel, nonparallel, or lines that coincide, depending upon their coefficients.
By using some analytic geometry you can transform a differential equation with nonparallel linear coefficients into a homogeneous form that you can then can solve using the methods used in the last two examples. When the coefficients describe parallel lines a simple substitution will transform the equation into a separable one immediately, that you can solve with the method demonstrated in Example 3.
Notice that a linear function that contains a constant will not be homogeneous. To remedy this, you determine the intersection point of the two lines, by equating them to zero, and solving the system. You then translate the origin to that new point. By determining the equations with respect to this new origin the constants drop out allowing you to solve the differential equation by the methods demonstrated in the last section. It is now simply a matter of substituting the equations of translation back in, for the final answer.
Expl #9 Non-Parallel Coefficients
Notice how the coefficients are linear and that they are not homogeneous (because of the constants). When you equate both to zero and solve for y, you can plot them in a LiveMath graph theory. Observe that they represent straight lines that intersect at a single point. By translating the axes to this point, you eliminate the constants. This, in turn, changes the two equations into a homogeneous form.
The following diagram shows the notation used to describe a translation from (0,0) to a new origin (h,k).
Since the coefficients describe two lines that intersect, by translating the origin using these equations, a new set of coefficient equations are created which are homogeneous. Again, this is because the new equations will not have constants in them.
The first thing you need to do is to find the coordinates of the new axes. You do this by finding the point of intersection of the two coefficient equations. Below LiveMath's matrix Op is used to solve the system.
The point of intersection (3,-4) now becomes the new origin by using these numbers to define h and k.
Notice how you can optain the new coefficients through inspection by merely dropping the constants.
The equation is now homogeneous and you can solve it using the methods found in the last two examples.
So that the manipulations will not be hard to follow, you may want to re-input the equation, without the subscripts, before you solve it. Once solved, re-label the x's and y's and substitute the translation equations back in for the final answer. The manipulations give you an implicit solution.
If the lines defined by the coefficients of dx and dy are parallel you can use a simple substitution which will leave the equation in a form which is separable. The method is similar to what you have done in the past. The key is to manipulate the equation by making a substitution that eliminates one of the variables.
Expl #10 Parallel Coefficients
NOTE: You can perform the same operation on the coefficient of dy to eliminate the y's.
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You can see, by observation, that you can place the coefficient of dy in terms of u. First note that by taking u times 3, an equation is generated that is close to this coefficient. By adding a constant to this equation you can complete the equality.
With Auto-ReManipulation on, change the 1 to a 5. This gives you the required equation.
Since lines that coincide are directly proportional to each other, coefficients that are defined by them are solved very easily with one substitution. For example, if the previous problem is altered in such a way that the coefficients are the same line (they are proportional), by dividing by the coefficient of dx you will have the following separable equation whose solution is easily obtained:
First Order Ordinary
