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**Unformatted text preview: **6 Simplifying Through Substitution In previous chapters, we saw how certain types of first-order differential equations (directly integrable, separable, and linear equations) can be identified and put into forms that can be integrated with relative ease. In this chapter, we will see that, sometimes, we can start with a differential equation that is not one of these desirable types, and construct a corresponding separable or linear equation whose solution can then be used to construct the solution to the original differential equation. 6.1 Basic Notions There are many first-order differential equations, such as dy dx = ( x + y ) 2 , that are neither linear nor separable, and which do not yield up their solutions by direct application of the methods developed thus far. One way of attempting to deal with such equations is to replace y with a cleverly chosen formula of x and “ u ” where u denotes another unknown function of x . This results in a new differential equation with u being the function of interest. If the substitution truly is clever, then this new differential equation will be separable or linear (or, maybe, even directly integrable), and can be be solved for u in terms of x using methods discussed in previous chapters. Then the function of real interest, y , can be determined from the original ‘clever’ formula relating u , y and x . Here are the basic steps to this approach, described in a little more detail and illustrated by being used to solve the above differential equation: 1. Identify what is hoped will be good formula of x and u for y , y = F ( x , u ) . This ‘good formula’ is our substitution for y . Here, u represents another unknown function of x (so “ u = u ( x ) ”), and the above equation tells us how the two unknown functions y and u are related. (Identifying that ‘good formula’ is the tricky part. We’ll discuss that further in a little bit.) 117 118 Simplifying Through Substitution Let’s try a substitution that reduces the right side of our differential equation, dy dx = ( x + y ) 2 , to u 2 . This means setting u = x + y . Solving this for y gives our substitu- tion,‘ y = u − x . 2. Replace every occurrence of y in the given differential equation with that formula of x and u , including the y in the derivative . Keep in mind that u is a function of x , so the dy / dx will become a formula of x , u , and du / dx (it may be wise to first compute dy / dx separately). Since we are using y = u − x (equivalently, u = x + y ), we have ( x + y ) 2 = u 2 , and dy dx = d dx [ u − x ] = du dx − dx dx = du dx − 1 . So, under the substitution y = u − x , dy dx = ( x + y ) 2 becomes du dx − 1 = u 2 . 3. Solve the resulting differential equation for u (don’t forget the constant solutions!). If possible, get an explicit solution for u in terms of x . (This assumes, of course, that the differential equation for u is one we can solve. If it isn’t, then our substitution wasn’t that clever, and we may have to try something else.)that clever, and we may have to try something else....

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