MATH 129 lesson 10-12

 



    First Order but higher degree differential equation:

A differential equation involving first order but degree higher than one is called first order but higher degree differential equation.

    Example: \( P^n+A_1 P^{n-1}+A_2 P^{n-2}+\ldots +A_{n-a} P+ A_n\)

is a first-order and n-th degree differential equation, where  \(p=\frac{d y}{d x}\) and \(A_1, A_2, \ldots , A_n\)

are functions of x,y


Convenient of solution and various types of form, The equation is solved by five methods,

  1. Solvable for p
  2. Solvable for y 
  3. solvable for x
  4. Clairauit's equation
  5. Lagrange's equation

    Solvable for p:
If the equation \(P^n+A_1 P^{n-1}+A_2 P^{n-2}+\ldots +A_{n-a} P+ A_n=0\) can be expressed into n-factors of the first order and first degree \((p-\alpha_1) (p-\alpha_2)(p-\alpha_3)\ldots (p-\alpha_n)=0 \) then equating to zero each factor, n-equation of the first order and first degree are obtained, Then the solutions are obtained by integration.
    

    Question: Solve,  \(p^2+2xp-3x^2=0 \)
    Solution: Given equation is, 
\begin{align*} & p^2+2xp-3x^2=0 \tag{1}\label{eq:1} \\ or, ~& p^2+3xp-xp-3x^2=0\\ or, ~& p(p+3x)-x(p+3x)=0 \\ or, ~& (p-x)(p + 3x)= 0\\ ~& \\ \therefore & ~ p - x = 0 \tag{2}\label{eq:2} \\ & ~ p + 3x = 0 \tag{3}\label{eq:3} \\ \end{align*}
From (2) we get, \begin{align*} & \frac{dy}{dx} - x = 0\\ & dy - x dx = 0 \end{align*}
Integrating this we get, \begin{align*} & y-\frac{x ^ 2 }{2}- \frac{c}{2}= 0 \\ or, ~ & 2y-x ^2 -c = 0 \end{align*}
From (3) we get, \begin{align*} & \frac{dy}{dx} +3x =0 \\ or,~ & dy + 3x~dx = 0 \end{align*}
Integrating this we get, \begin{align*} & y + \frac{3x2 ^2}{2} - \frac{c}{2}= 0\\ & 2y + 3x^2 - c = 0 \end{align*}
General solution of (1) is, \begin{align*} (2y - x^2 - c )(2y + 3x^2 - c ) = 0 \end{align*}


Post a Comment

Post a Comment (0)

Previous Post Next Post