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: y'=\frac{t^2+3y^2}{2ty}.

Solutions: As we can see, this is not a separable equations. However, it is Euler Homogeneous, since f(ct,cy)=f(t,y). Thus we can convert it to: \frac{1+3(\frac{y}{t})^2}{2 (\frac{y}{t})} which is the desired form of y'=F(\frac{y}{t}). Hence, we have to transform the equation by v=\frac{y}{t} and arrive at the equation: \frac{v'}{F(v)-v}=\frac{1}{t}, which is a separable equation and is easily solvable. By following the steps shown in earlier parts, we arrive at the final solution structure: y(t)=^+_-t\sqrt{c_1t-1}, for some constant c_1 \in \mathbb{R}.

Exercise 2: Find all solutions of 2tyy'+2t+y^2=0.

Solution: As given in previous article, we can see that here we have to first calculate the potential \psi such that N=\frac{\partial \psi}{\partial y} and M=\frac{\partial \psi}{\partial t}, where N(t,y)=2ty and M(t,y)=2t+y^2. After some very trivial calculations, we get the requires potential to be: \psi(t,y)=ty^2+t^2, and hence the required solutions is: ty^2+t^2=c for some constant c.

 

Exercise 3: Find all solutions of (t^2+ty)y'+3ty+y^2=0.

Solutions: Here we can see:
N(t,y)=t^2+ty \Rightarrow \frac{\partial N}{\partial t}=2t+y

M(t,y)=3ty+y^2 \Rightarrow \frac{\partial M}{\partial y}=3t+2y

Thus the equation is not exact. However, h=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t} }{N}  depends only on t and not on y, in fact it equals to \frac{1}{t}. Thus by Theorem 6.2 we conclude that this equation is semi-exact,and (e^HN)y'+e^HM=0 is exact, where H is the anti-derivative of h. Thus we just use the equation: t(t^2+ty)y'+t(3ty+y^2)=0 and solve it just like the above exact equation.

 

Exercise 4: Find all solutions for (5te^{-y}+2cos(3t))y'+(5e^{-y}-3sin(3t))=0

Solution:   We can easily check the equation is not exact: \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}M \neq 0. Calculating l=-\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t}}{M}  gives the value of l to be 2. Hence, (e^LM)t'+e^LN=0 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 y'(t)=f(t,y(t)) 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: y'(t)=f(t,y(t)), \textrm{ }y(t_0)=y_0If the function f is continuous in the domain D_a=[t_0-a,t_0+a] \times [y_0-a,y_0+a] \subset \mathbb{R}^2 for some a \gt 0, and f is Lipschitz Continuous on y, then there exists a positive b \lt a such that there exists an unique solution of y, on the domain [t_0-b,t_0+b] , to the initial value problem.

Overview of Solution:

  • Step 1: \int_{t_0}^t y'(s)ds=\int_{t_0}^t f(s,y(s))ds \Rightarrow y(t)=y_0+\int_{t_0}^t f(s,y(s))ds
  • Step 2: Construct the sequence \left \{ y_n \right \} such that y_{n+1}(t)=y_0+\int_{t_0}^t f(s,y_n(s))ds, \textrm{ } n \geq 0 \textrm{ } y_0(t)=y_0 This is called Picard’s Iteration.
  • Step 3: Show that \left \{ y_n \right \} is a Cauchy Sequence in the space of uniformly continuous functions.
    [Hint: Check that ||y_{n+1}-y_n|| \leq r^n||y_n-y_{n-1}|| \textrm{, } r \lt 1]
  • Step 4: The proof of uniqueness follows similar steps from the Cauchy Proof.

Exercise: Use Picard’s Iteration to solve y'=2t^4y.  (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: y'(t)=\frac{t^2}{y^4(t)+a_3y^3(t)+a_2y^2(t)+a_1y(t)+a_0}.

Proceeding as suggested earlier by the Picard’s Iteration Process, we get the solution as: \frac{y^5(t)}{5}+\frac{a_3 y^4(t)}{4t}+\frac{a_2y^3}{3t}+\frac{a_1y^2}{2t}+a_1y(t)=\frac{t^3}{3}+c.

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: y'(t)=y^{\frac{1}{3}}(t) \textrm{, }y(0)=0

Now the equation has two solutions, one is y(t)=0. The other solution can be found using the method of solution for Separable Equation.

Property 3: The domain of a solution y  to a nonlinear initial value problem may change when we change the initial data y_0.

Example: Find all solutions of: y'(t)=y^2(t) \textrm{, }y(0)=y_0.

The solution is: y(t)=\frac{1}{\frac{1}{y_0}-t}. The solution diverges at t=\frac{1}{y_0}  so the domain of the solution y is not the whole real line \mathbb{R}. Instead, the domain is \mathbb{R}-\left \{ y_0 \right \}, so it depends on the values of the initial data y_0.

 

That’s it for today. I will come back again with some new updates on this topic. Till Then, Goodbye.


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