In my last posts: Ordinary Differential Equations: Part-1 and Part-2, we have learnt the methods to solve certain types of differential equations – Linear with constant and variable coefficients, Bernoulli Equations, Separable Equations, Euler Homogeneous Equations, Exact and Semi-Exact Differential Equations. Before jumping on the solution methods for some additional types of differential equations, let us first take care of some loose ends from my last article: Ordinary Differential Equations: Part-2.
Exercise 1: Give all solutions of: .
Solutions: As we can see, this is not a separable equations. However, it is Euler Homogeneous, since . Thus we can convert it to: which is the desired form of . Hence, we have to transform the equation by and arrive at the equation: , which is a separable equation and is easily solvable. By following the steps shown in earlier parts, we arrive at the final solution structure: , for some constant .
Exercise 2: Find all solutions of .
Solution: As given in previous article, we can see that here we have to first calculate the potential such that and , where and . After some very trivial calculations, we get the requires potential to be: , and hence the required solutions is: for some constant c.
Exercise 3: Find all solutions of .
Solutions: Here we can see:
Thus the equation is not exact. However, depends only on t and not on y, in fact it equals to . Thus by Theorem 6.2 we conclude that this equation is semi-exact,and is exact, where H is the anti-derivative of h. Thus we just use the equation: and solve it just like the above exact equation.
Exercise 4: Find all solutions for
Solution: We can easily check the equation is not exact: , . Calculating gives the value of l to be 2. Hence, is the required exact equation, where L is the anti-derivative of l. (Finish the exercise).
Solving non-linear equations is infinite times harder than linear equations. We have discussed some methods of solving certain types of non-linear equations, but that doesn’t even scratch the surface. It is more prudent at this point to turn our attention to determine which non-linear equations are actually solvable, which brings us to the Picard-Lindelöf Theorem, which doesn’t give a general formula for solving non-linear equations.
Definiton: An ordinary Differential Equation is non-linear iff the functionf is non-linear in it’s second argument.
Theorem 8.1: (Picard-Lindelöf Theorem): Consider the initial value problem: If the function f is continuous in the domain for some , and f is Lipschitz Continuous on y, then there exists a positive such that there exists an unique solution of y, on the domain , to the initial value problem.
Overview of Solution:
- Step 1:
- Step 2: Construct the sequence such that This is called Picard’s Iteration.
- Step 3: Show that is a Cauchy Sequence in the space of uniformly continuous functions.
[Hint: Check that ]
- Step 4: The proof of uniqueness follows similar steps from the Cauchy Proof.
Exercise: Use Picard’s Iteration to solve . (Solution will be given in Part 4)
Properties of Non-Linear Differential Equation:
Let’s Discuss some properties of non-linear equations with examples.
Property 1: There is no explicit formula for the solution to every nonlinear differential equation.
Proceeding as suggested earlier by the Picard’s Iteration Process, we get the solution as: .
The solution is the root of a polynomial degree five for all possible values of the polynomial coefficients. But it has been proven that there is no formula for the roots of a general polynomial degree bigger or equal five. Thus there is no explicit expression for solutions of the equation.
Property 2: Solutions to initial value problems for nonlinear equations may be non-unique when the function f does not satisfy the Lipschitz condition.
Now the equation has two solutions, one is . The other solution can be found using the method of solution for Separable Equation.
Property 3: The domain of a solution to a nonlinear initial value problem may change when we change the initial data .
Example: Find all solutions of: .
The solution is: . The solution diverges at so the domain of the solution y is not the whole real line . Instead, the domain is , so it depends on the values of the initial data .
That’s it for today. I will come back again with some new updates on this topic. Till Then, Goodbye.