# Ordinary Differential Equations: Part-3 1

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).

## Today’s Lessons:

### Non-Linear Equations:

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.

Example: .

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.

Example:

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.