Ordinary Differential Equations: Part 2 1


In previous article: Ordinary Differential Equations: Part I, we have learnt the basics of Ordinary Differential Equations, and solution methods of few particular types of Differential Equations. Let’s make a quick review with a few example:

Quick Review:

  • Example 1: Find all solutions to the constant coefficient equation: y'=6y+9.

 Solution: Using Theorem 1 from the previous article, we can give the result to be:

y(t)=ce^{at}-\frac{b}{a}=ce^{6t}-\frac{9}{6}=ce^{6t}-\frac{3}{2}

  • Example 2: Find all solutions to the variable coefficient equation: ty'=-2y+4t^2.

Solution: Using Theorem 2 of previous article, we will solve this problem.

    • Thus, a(t)=-\frac{2}{t} and b(t)=4t.
    • A(t)=\int a(t)dt=\int -\frac{2}{t}dt=-2ln(t)
    • y(t)=ce^{A(t)}+e^{A(t)}\int e^{-A(t)}b(t)dt=ct^{-2}+t^{-2}\int \frac{4t}{t^{-2}}dt
    • Solving the above integral we get: y(t)=\frac{c}{t^2}+t^2. Thus solved.
  • Example 3: Find all solutions of y'=ay+by^4

Solution: Here we will use Theorem 3 of last article, the theorem for Bernoulli Equations:

Hence, we have to transform this equation into a linear equation at first. This is done by the transformation: v=\frac{1}{y^{n-1}}=\frac{1}{y^3}. Then the equation transforms to:

The solution of this linear equation with constant coefficients is: v=e^{-3at}-\frac{b}{a}.

Thus the final solution becomes: \frac{1}{y^3}=e^{3at}-\frac{b}{a}  thereby giving: y(t)=(e^{3at}-\frac{b}{a})^{-3}

  • Example 4: Solving the following separable differential equation: y'=-y^2cos(2t)

Solution: I will use Theorem 4 of last article. Here h(y)=-\frac{1}{y^2} and g(t)=cos(2t). Hence we have to find anti-derivatives H(t) and G(t) of h(t) and g(t) respectively. Doing that, we get: H(y)=\frac{1}{y} and G(t)=\frac{sin(2t)}{2}. Hence we get the solution: y(t)=\frac{2}{sin(2t)+2c}

Today’s Lessons:

Euler Homogeneous Equations:  

Now till now, we have learnt how to solve separable differential equations. However, in many cases, the equations might not be represented as separable equations. However, Euler Homogenous Differential Equations, gives a set of equations, which after a proper transformation becomes separable, and hence can be solved using the above procedure.

Example: y'=\frac{t^2+3y^2}{2ty}. Obviously this cannot be represented in a separable form, so we cannot apply Theorem 4 of previous article. However, after a simple transformation this can be converted to a separable equation, and then can be solved easily.

Definition: An Euler homogeneous differential equation has the form: y'(t)=F(\frac{y(t)}{t})

Thus, right hand side is scale-invariant, and hence particular case of homogeneous functions of degree n, which are functions f satisfying: f(ct,cy)=c^nf(t,y).

Theorem 5.1: If the functions N, M, of t, y, are homogeneous of the same degree, then the differential equation N(t,y)y'(t)+M(t,y)=0 is Euler homogeneous.

Theorem 5.2: The Euler Homogeneous Equation y'(t)=F(\frac{y(t)}{t})  for the function y determines a separable differential equation for v=\frac{y}{t} given by \frac{v'}{F(v)-v}=\frac{1}{t}.

Thus, we have got a method to solve for Euler Homogeneous Equations.

Exact Differential Equation:

Definition:  An exact differential equation for y is: N(t,y)y'+M(t,y)=0, where the functions N and M satisfies \frac{\partial}{\partial t}N(t,y)=\frac{\partial}{\partial y}M(t,y).

It is easy to verify that separable equations are always exact, however the reverse may not be always true. Also, linear equations are not exact.

Poincaré Theorem: Continuously differentiable functions M, N, on t, y satisfies \frac{\partial N(t,y)}{\partial t}=\frac{\partial M(t,y)}{\partial y} iff there is a twice continuously differentiable function \psi(t,y) such that \frac{\partial \psi(t,y)}{\partial y}=N(t,y) and \frac{\partial \psi(t,y)}{\partial t}=M(t,y).

Theorem 6.1: If the differential equation N(t,y)y'+M(t,y)=0 is exact, then it can be written as  \frac{d \psi}{dt}(t,y(t))=0, where \psi is a potential function and satisfies N=\frac{\partial \psi}{\partial y} and M=\frac{\partial \psi}{\partial t}. Hence solutions of the exact equations can be given in implicit form: \psi(t,y(t))=c \textrm{, } c \in \mathbb{R}.

Definition: A semi-exact differential equation is a non-exact equation that can be transformed into an exact equation after a multiplication by an integrating factor.

Example:  Linear Differential equations are semi-exact. (Prove)

Theorem 6.2: If the equations: N(t,y)y'+M(t,y)=0 is not exact, with \frac{\partial N}{\partial t} \neq \frac{\partial M}{\partial y}, the function N \neq 0 and where the function h is defined as h=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t} }{N} depends only on t, not on y, then the equation below is exact. (e^HN)y'+e^HM=0, where H is an anti-derivative of h.

Note: The function \mu(t)=e^{H(t)} is called an integrating factor.

Example: Find all solutions of (t^2+ty)y'+3ty+y^2=0. (Try it. The Solution will be updated in the Part-III)

Inverse Functions:

Sometimes equations for y is not exact or semi-exact, but that of y^{-1} might be. Suppose t(y) is the inverse of y(t), giving: t'(y)=\frac{1}{y'(t)}.

Theorem 7.1: Ny’+M=0 is exact \Leftrightarrow Mt’+N=0 is exact.

Hence, for exact equations, inversion has zero effect.

Theorem 7.2: If Mt’+N=0 is not exact, with \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}M \neq 0 and where l is defined as l=-\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t}}{M} depends only on y, not on t, then the following equation is exact: (e^LM)t'+e^LN=0, where L is anti-derivative of l.

Note: The function \mu(y)=e^{{L(y)} is called an integrating factor.

Exercise: Try to solve: (5te^{-y}+2cos(3t))y'+(5e^{-y}-3sin(3t))=0 (Solution will be given in Part III)

That’s all for today. Try to solve some differential equations of the aforementioned types to get comfortable with the methods. My next post will be on Non-Linear Equations, and Picard-Lindelöf Theorem. Till then, Goodbye.

 

 

 


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